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Page No 487:

Question 1:

Show that the progression 3, 9, 15, 21,... is an AP. Write down its first term and the common difference.

Answer:

The given progression is 3, 9, 15, 21,...

Clearly, (9 - 3) = (15 - 9) = (21 - 15) = 6 (Constant)
Thus, each term differs from its preceding term by 6.
So, the given progression is an AP.
Its first term is 3 and the common difference is 6.
Thus, a = 3, and d = 6
 

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Question 2:

Show that the progression 16, 11, 6, 1, −4,... is an AP.
Write down its first term and the common difference.

Answer:

The given progression is 16, 11, 6, 1, -4,...
Clearly, (11 - 16) = (6 - 11) = (1 - 6) = (-4 - 1) = -5 (Constant)

Thus, each term differs from its preceding term by -5.
So, the given progression is an AP.
Its first term, a is 16 and common difference, d is -5.

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Question 3:

Find:
(i) The 20th term of the AP 1, 5, 9. 13, 17,...
(ii) The 35th term of the AP 6, 9, 12, 15,...
(iii) The nth term of the AP 5, 11, 17, 23,...
(iv) The 11th term of the AP (5ax), 6a, (7a + x),....

Answer:

(i) The given AP is  1, 5, 9, 13, 17,...
    First term, a = 1 and common difference, d = (5 - 1) = 4
    T20 = a + (20 - 1)d = a + 19d

          = 1 + 19 ⨯ 4 = 77   
   ∴ 20th term  = 77

(ii)  T35 = a + (35 - 1)d = a + 34d
            = 6 + 34 ⨯ 3 = 108                                    [∵ a = 6 and d = 3]
   ∴ 35th term = 108

(iii) Tn = a + (n - 1)d  
           = 5 + (n - 1) ⨯ 6
           = 5 + 6n - 6 = 6n - 1                 [∵ a = 5 and d = 6]
      ∴  nth term = 6n - 1

(iv) T11 = a + (11 - 1)d = a + 10d
           = 5a - x + 10 ⨯ (a + x)                                       [∵ a = (5a -x) and d = {6a -(5a - x)} = a + x]
           =  5a - x + 10a + 10x = 15a + 9x
​         ∴ 11th term = 15a + 9x
​    

Page No 487:

Question 4:

Find:
(i) The 10th term of the AP 63, 58, 53, 48,...
(ii) The 14th term of the AP 9, 5, 1, −3,...
(iii) The nth term of the AP 16, 9, 2, −5,...

Answer:

(i) The given AP is 63, 58, 53,48,...
    First term, a = 63 and common difference, d = (58 - 63) = -5
    T10 a + (10 - 1)d = a + 9d

         = 63 + 9 ⨯ (-5) = 63 - 45 = 18
   ∴ 10th term = 18

(ii)  T14 = a + (14 - 1)d = a + 13d
            = 9 + 13 ⨯ (-4) = 9 - 52 = -43                                    [∵ a = 9 and d = - 4]
     ∴ 14th term  = - 43

(iii) Tn = a + (n - 1)d  
           = 16 + (n - 1) ⨯ (-7)          [∵ a = 16 and d = - 7]
           = 16 - 7n + 7 = 23 - 7n                
     ∴ nth term = 23 -7n

 

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Question 5:

Find the 37th term of the AP 6,734, 912, 1114,...

Answer:

The given AP is 6, 734, 912, 1114,... 
First term, a = 6 and common difference, d = 734 - 6  314 -6  31 - 244 = 74
Now, T37 = a+(37-1) d = a+36d       =6+36×74=6+63=69 37th term=69

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Question 6:

Find the 25th term of the AP 5, 412, 4, 312, 3,...

Answer:

   The given AP is 5, 412, 4, 312, 3,... 
    First term = 5
    Common difference= 412 - 5  92 -5  9 - 102 = -12
     a =5 and d=-12Now, T25 = a+(25-1)d=a+24d       =5+24×-12=5-12=-7 25th term=-7

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Question 7:

How many terms are there in the AP 6, 10, 14, 18,...,174?

Answer:

In the given AP, a  = 6 and d = (10 - 6) = 4
Suppose that there are n terms in the given AP.
Then Tn = 174
a + (n - 1)d = 174
⇒ 6 + (n - 1) ⨯ 4 = 174
⇒ 2 + 4n = 174
⇒ 4n = 172
n = 43
Hence, there are 43 terms in the given AP.
             

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Question 8:

How many terms are there in the AP 41, 38, 35,...., 8 ?

Answer:

In the given AP, a  = 41 and d = (38 - 41) = -3
Suppose that there are n terms in the given AP.
Then Tn = 8
⇒ a + (n - 1) d = 8
⇒ 41 + (n - 1) ⨯ (-3) = 8
⇒ 44 - 3n = 8
⇒ 3n = 36
n = 12
Hence, there are 12 terms in the given AP.

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Question 9:

Which term of the AP 3, 8, 13, 18,... is 88?

Answer:

In the given AP, first term, a = 3 and common difference, d = (8 - 3) = 5.

Let's its nth term be 88.
Then Tn = 88
⇒ a + (n - 1)d = 88
⇒ 3 + (n - 1) ⨯ 5 = 88
⇒ 5n - 2 = 88
⇒ 5n = 90
n = 18
Hence, the 18th term of the given AP is 88.

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Question 10:

Which term of the AP 72, 68, 64, 60,... is 0?

Answer:

In the given AP, first term, a = 72 and common difference, d = (68 - 72) = -4.
Let its nth term be 0.
Then, Tn = 0  
⇒ a + (n - 1)d = 0
⇒ 72 + (n - 1) ⨯ (-4) = 0
⇒ 76 - 4n = 0
⇒ 4n = 76
n = 19
Hence, the 19th term of the given AP is 0.

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Question 11:

Which term of the AP 56, 1, 116,113,.... is 3?

Answer:

In the given AP, first term = 56 and common difference, d =1- 56 = 16.
Let its nth term be 3.

Now, Tn=3    a + (n-1)d = 356 + (n-1)×16 = 323 + n6 = 3n6 = 73n = 14          

Hence, the 14th term of the given AP is 3.

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Question 12:

Find the value of x for which (5x + 2), (4x − 1) and (x + 2) are in AP.

Answer:

Let the terms (5x +2), (4x -1) and (x +2) form an AP.

∴ (4x -1) - (5x +2) = (x +2) - (4x -1)             [ ∵ Common difference remains constant]
⇒ -x - 3 = -3x + 3
⇒ 2x = 6
x = 3

Hence, when x = 3, the given terms are in AP.

Page No 487:

Question 13:

If the nth term of a progression is (4n − 10), show that it is an AP. Find its
(i) first term, (ii) common difference, and (iii) 16th term.

Answer:

Tn = (4n - 10)     [Given]
​T1 = (4 ⨯ 1 - 10) = -6
T2 = (4 ⨯ 2 - 10) = -2
T3 = (4 ⨯ 3 - 10) = 2
T4 = (4 ⨯ 4 - 10) = 6

Clearly, [ -2 - (-6)] = [2 - (-2)] = [6 - 2] = 4               (Constant)
So, the terms -6, -2, 2, 6,... forms an AP.

Thus, we have;
(i) First term = -6
(ii) Common difference = 4
(iii) T16 = a + (n -1)d  = a + 15d = ​- 6 + 15 ⨯ 4 = 54

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Question 14:

The 4th and 10th terms of an AP are 13 and 25 respectively. Find the first term and the common difference of the AP. Also, find its 17th term.

Answer:

We have;
T4 = a + (n - 1)d 
⇒ a + 3d = 13          ...(1)

​T10 = a + (n - 1)d 
⇒ a + 9d = 25           ...(2)
On solving (1) and (2), we get:
a = 7 and d = 2


Thus, we have the following:
(i) First term = 7
(ii) Common difference = 2
(iii) T17 = a + (n - 1)d  = a + 16d = ​7 + 16 ⨯ 2 = 39

Page No 487:

Question 15:

The 8th term of an AP is 37 and its 12th term is 57. Find the AP.

Answer:

We have:
T8 = a + (n -1)d  
⇒ a + 7d = 37             ...(1)

​T12 = a + (n -1)d  
⇒ a + 11d = 57            ...(2)
On solving (1) and (2), we get:
a = 2 and d = 5

Thus, first term = 2 and common difference = 5

∴ The terms of the AP are 2, 7, 12, 17,...



Page No 488:

Question 16:

The 7th term of an AP is −4 and its 13th term is −6. Find the AP.

Answer:

We have:
T7 = a + (n - 1)d 
⇒ a + 6d = -4           ...(1)

​T13 = a + (n - 1)d  
⇒ a + 12d = -16          ...(2)
On solving (1) and (2), we get:
a = 8 and d = -2


Thus, first term = 8 and common difference = -2

∴ The terms of the AP are 8, 6, 4, 2,...

Page No 488:

Question 17:

If the 10th term of an AP is 52 and the 17th term is 20 more than the 13th term, find the AP.

Answer:

In the given AP, let the first term be a and the common difference be d.
Then, Tna + (n - 1)d ​
Now, we have:
T10 = a + (10 - 1)d 
⇒ a + 9d  = 52       ...(1)
T13a + (13 - 1)d = a + 12d               ...(2)
T17 = a + (17 - 1)d = a + 16d                 ...(3)     

But, it is given that T17 = 20 + T13
i.e., a + 16d  = 20 + a + 12d 
⇒ 4d = 20  
d = 5
On substituting d = 5 in (1), we get:
a + 9 ⨯ 5 = 52 
⇒​ a = 7

Thus, a = 7 and d = 5

∴ The terms of the AP are 7, 12, 17, 22,...

Page No 488:

Question 18:

The 4th term of an AP is zero. Prove that its 25 term is triple is 11th term.

Answer:

In the given AP, let the first term be a and the common difference be d.
Then, Tn = a + (n - 1)d ​
Now, T4 = a + (4 - 1)d 
⇒ a + 3d  = 0        ...(1)

⇒ a = -3d  

Again, T11 = a + (11 - 1)d  = a + 10d
=​ -3d + 10 d = 7d             [ Using (1)]

Also, T25 = a + (25 - 1)d =​ a + 24d = -3d + 24d = 21d             [ Using (1)]​
    i.e., T25 = 3 ⨯ 7d = (3 ⨯ T11)
​Hence, 25th term is triple its 11th term.

Page No 488:

Question 19:

Which term of the AP 3, 8, 13, 18, ... will be 55 more than its 20th term?

Answer:

Here, a = 3 and d = (8 - 3) = 5
The 20th term is given by
T20 = a + (20 - 1)d = a + 19d = 3 + 19 ⨯ 5 = 98
∴ Required term = (98 + 55) = 153
Let this be the nth term.
Then Tn = 153
⇒ 3 + (n - 1) ⨯ 5 = 153
⇒ 5n = 155
n = 31
Hence, the 31st term will be 55 more than its 20th term.  

Page No 488:

Question 20:

Which term of the AP 5, 15, 25, ...will be 130 more than its 31st term?

Answer:

Here, a = 5 and d = (15 - 5) = 10
The 31st term is given by
T31 = a + (31 - 1)d = a + 30d = 5 + 30 ⨯ 10 = 305
∴ Required term = (305 + 130) = 435
Let this be be the nth term.
Then Tn = 435
​⇒ 5 + (n - 1) ⨯ 10 = 435
⇒ 10n = 440
n = 44
Hence, the 44th term will be 130 more than its 31st term.  

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Question 21:

For what value of n, the nth terms of the arithmetic progressions 63, 65, 67,... and 3, 10, 17,... are equal?

Answer:

Let the nth term of the given progressions be tn and Tn, respectively.
The first AP is 63, 65, 67,...
Let its first term be a and common difference be d.
Then a = 63 and d = (65 - 63) = 2
So, its nth term is given by
tn = a + (n - 1)
63 + (n - 1) ⨯ 2
61 + 2n

The second AP is 3, 10, 17,...
Let its first term be A and common difference be D.
Then A = 3 and D = (10 - 3) = 7
So, its nth term is given by
Tn = A + (n - 1)
 3 + (n - 1) ⨯ 7
⇒ 7n - 4
Nowtn = ​Tn
 
⇒ 61 + 2n= 7n - 4​
65 = 5n
n = 13
Hence, the 13th terms of the AP's are the same.

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Question 22:

How many 3-digit numbers are divisible by 7?

Answer:

The smallest three-digit number which is divisible by 7 is 105 and the largest three-digit number divisible by 7 is 994.
Hence, the progression is 105, 112, 119,..., 994
Here, a = 105, d = 7 and last term = 994
Let n be the number of terms of the sequence.
Now, we have:
Tn = 994 a + (n - 1)d = 994
              ⇒ 105 + (n - 1) ⨯ 7 = 994
              7n-7 = 889
              ⇒ 7n = 896
              ⇒ n = 128

Hence, there are 128 three-digit numbers which are divisible by 7.

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Question 23:

Find the 8th term from the end of the AP 7, 10, 13, ..., 184.

Answer:

Here, a = 7 and d = (10 - 7) = 3, l = 184 and n = 8th from the end.
Now, nth term from the end = [l - (n -1)d]
8th term from the end = [184 - (8 - 1) ⨯ 3]
                                 = [184 - (7 ⨯ 3)] = (184 - 21) = 163
Hence, the 8th term from the end is 163.

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Question 24:

Find the 6th term from the end of the AP 17, 14, 11, ..., (−40).

Answer:

Here, a = 17 and d = (14 - 17) = -3, l = (-40) and n = 6
Now, nth term from the end = [l - (n - 1)d]  
6th term from the end = [(-40) - (6 - 1) ⨯ (-3)]
                                 = [-40 + (5 ⨯ 3)] = (-40 + 15) = -25
Hence, the 6th term from the end is -25.

Page No 488:

Question 25:

Find the 11th term from the last term towards the first term of the AP 10, 7, 4,.... (−62).

Answer:

Here, a = 10 and d = (7 - 10) = -3, l = (-62) and n = 11
Now, nth term from the end = [l - (n - 1)d]  
11th term from the end = [(-62) - (11 - 1) ⨯ (-3)]
                                   = [-62 + (10 ⨯ 3)] = (-62 + 30) = -32

Hence, the 11th term from the last towards the first term of the AP is -32.

Page No 488:

Question 26:

If the pth term of an AP is q and its qth term is p then show that its (p + q)th term is zero.

Answer:

In the given AP, let the first term be a and the common difference be d.
Then Tn = a + (n - 1)d ​
⇒ Tp = a + (p - 1)d = q               ...(i)
     ​

 ⇒ Tq = a + (q - 1)d = p              ...(ii)   

On subtracting (i) from (ii), we get:
(q - p)d = (p - q)
 ⇒ d = -1
Putting d = -1 in (i), we get:
a = (pq - 1)

Thus, a = (p + q - 1) and d = -1
 Now, Tp+q = a + (pq - 1)d
                  =​ (p + q - 1) + (p + q - 1)(-1) 
                  = (p + q - 1) - (p + q - 1) = 0 ​
                                              
 ​Hence, the (p+q)th term is 0 (zero).

Page No 488:

Question 27:

If 10 times the 10th term of an AP is equal to 15 times the 15th term, show that its 25th term is zero.

Answer:

In the given AP, let the first term be a and the common difference be d.
Now, Tn = a + (n - 1)d ​
Then we have:
T10 = a + (10 - 1)d = a + 9d   

T15 = a + (15 - 1)d  = a + 14d               

Now, (10 ⨯ T10) = (15 ⨯ T15
⇒10[ a + 9d]  = 15 [a + 14d ]
⇒ 10a + 90d = 15a + 210d
⇒ -5a = 120d
a = -24d

∴ T25 = a + (25 - 1)d =​ a + 24d = -24d + 24d = 0             [ ∵ a = -24d]​
 ​Hence, the 25th term is 0 (zero).

Page No 488:

Question 28:

The first and last terms of an AP are a and l, respectively. Show that the sum of the nth term from the beginning and the nth term from the end is (a + 1).

Answer:

In the given AP, first term = a and last term = l.
Let the common difference be d.
Then, nth term from the beginning is given by
Tn = a + (n - 1)d             ...(1)

Similarly, nth term from the end is given by
Tn = {l - (n - 1)d}            ...(2)
Adding (1) and (2), we get:
a + (n - 1)d + {l - (n - 1)d}​
= a + (n - 1)d + l - (n - 1)d
= a
+ l 
                
​Hence, the sum of the nth term from the beginning and the nth term from the end is (a + l) .

Page No 488:

Question 29:

In a flower bed, there are 43 rose plants in the first row, 41 in the second, 39 in the third, and so on. There are 11 rose plants in the last row. How many rows are there in the flower bed?

Answer:

The numbers of rose plants in consecutive rows are 43, 41, 39,..., 11.

Difference of rose plants between two consecutive rows = (41 - 43) = (39 - 41) = -2  [Constant]
So, the given progression is an AP.
Here, first term = 43
Common difference = -2
Last term = 11
Let n be the last term, then we have:
Tn = a + (n - 1)d
⇒ 11 = 43 + (n - 1)(-2)
⇒ 11 = 45 - 2n
⇒ 34 = 2n
⇒ n = 17
Hence, the 17th term is 11 or there are 17 rows in the flower bed.

Page No 488:

Question 30:

A sum of Rs 1000 is invested at 8% per annum simple interest. Calculate the interest at the end of each year. Show that these interest form an AP. Using this fact, find the interest at the end of 30 years.

Answer:

Simple interest = P×R×T100 , where P=principal, R=rate of interest and T=time in years.                        = 1000×8×T100 = 80×T
On putting T = 1, 2, 3, 4, we get:
Rs 80, Rs 160, Rs 240, Rs 320 and so on.

The difference between the interests of two consecutive years = (160 - 80) = (240 - 160) = Rs 80 [Constant]
So, the progression of simple interests forms an AP.
Here, first term = Rs 80
Common difference = Rs 80

Now, interest at the end of 30 years, T30 = a + (n - 1)d
= 80 + (30 - 1)(80) = 80 + 29(80) = Rs 2400 
Hence, the interest at the end of 30 years is Rs 2400.
 



Page No 490:

Question 1:

Write the value of x for which (x + 2), 2x, (2x + 3) are three consecutive terms of an AP?

Answer:

Since (x + 2), 2x and (2x + 3) are in AP, we have:
2x - (x+2) = (2x+3-2x
⇒ x - 2 = 3
x = 5​
∴ x = 5

Page No 490:

Question 2:

Find three numbers in AP whose sum is 15 and product is 80.

Answer:

Let the required numbers be (a - d), a and (a + d).
Then (a - d) + a + (a + d) = 15
⇒ 3a = 15
⇒ a = 5​
Also, (a - d).a.(a + d)​ = 80
⇒ a(a2 - d2) = 80​   
5 ​(25 - d2) = 80​   ​  
d2 = 25​ - 16 =  9 
d = ±3
Thus, a = 5 and ±3
 Hence, the required numbers are (2, 5 and 8) or ( 8, 5 and 2).

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Question 3:

The sum of three numbers in AP is 27 and their product is 405. Find the numbers.

Answer:

Let the required numbers be (a - d), a and (a + d). 
Then (a - d) + a + (a + d) = 27
⇒ 3a = 27
⇒ a = 9​
Also, (a - d).a.(a + d)​ = 405
⇒ a(a2 - d2) = 405​   
⇒ 9​(81​​ - d2) = 405​ ​   ​  
d2 = 81​​ - 45 = 36 
d = ±6
Thus, a = 9 and ±6
 Hence, the required numbers are (3, 9 and 15) or (15, 9 and 3).

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Question 4:

The sum of three numbers in AP is 3 their product is −35. Find the numbers.

Answer:

Let the required numbers be (a - d), a and (a + d). 
Then (a - d) + a + (a + d) = 3
⇒ 3a = 3
⇒ a = 1​
Also, (a - d).a.(a + d)​ = -35
⇒ a(a2 - d2) = -35​  
1​.​(1​ - d2) = -35​ ​   ​  
d2 = 36 
d = ±6

Thus, a = 1 and ±6
 Hence, the required numbers are ( -5, 1 and  7) or ( 7, 1 and -5).

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Question 5:

Divide 24 in three parts such that they are in AP and their product is 440.

Answer:

Let the required parts of 24 be (a - d), a and (a + d) such that they are in AP. 
Then (a - d) + a + (a + d) = 24
⇒ 3a = 24  
⇒ a = 
8​
Also, (a - d).a.(a + d)​ = 440
⇒ a(a2 - d2) = 440​   
⇒ 8​(64​​ -d2) = 440​    ​  
d2 = 64 - 55 = 9 
 ±3
 Thus, a = 8 and ±3
 Hence, the required parts of 24 are (5, 8,11) or (11, 8, 5).

Page No 490:

Question 6:

The sum of three consecutive terms of an AP is 21 and the sum of the squares of these terms is 165. Find these terms.

Answer:

Let the required terms be (a - d), a and (a + d). 
Then (a - d) + a + (a + d) = 21
⇒ 3a = 21
⇒ a = 7​
Also, (a - d)2 + a2 + (a + d)2​ = 165
⇒ 3a2 + 2d2 = 165  
​(3 ⨯ 49​​ + 2 d2) = 165   ​  
⇒ 2d2 = 165 - 147 = 18 
 d2  = 9​
⇒ d = ±3
 Thus, a = 7 and ±3
 Hence, the required terms are ( 4, 7,10) or ( 10, 7, 4).

Page No 490:

Question 7:

The angle of a quadrilateral are in AP whose common difference is 10°. Find the angles.

Answer:

Let the required angles be (a - 15)o, (a - 5)o, (a + 5)o and (a + 15)o, as the common difference is 10 (given).
Then (a - 15)o + (a - 5)o + (a + 5)o + (a + 15)o = 360o
⇒ 4a = 360 
⇒ a = 
90

Hence, the required angles of a quadrilateral are 90-15°, 90-5°, 90+5° and 90+15°; or 75°, 85°, 95° and 105°.

Page No 490:

Question 8:

Find your numbers in AP whose sum is 28 and the sum of whose squares is 216.

Answer:

Let the required numbers be (a - 3d)(a - d), (a + d) and (a + 3d). 
Then (a - 3d) + (a - d)​ + (a + d) + (a + 3d) = 28
⇒ 4a = 28
⇒ a = 7​

Also, (a - 3d)2 + (a - d)+ (a + d)2​ + (a + 3d)2 = 216
⇒ 4 (​a2 + 5d2)​= 216  
4⨯​( 49​ + 5d2)​ = 216   ​  
⇒ 5d2 = 54 - 49 = 5 
 d21​
⇒ d = ±1
 Thus, a = 7 and ±1
 Hence, the required numbers are (4, 6, 8, 10) or (10, 8, 6, 4).



Page No 500:

Question 1:

Find the sum of first 19 terms of the AP 2, 7, 12, 17, ..... .

Answer:

Here, a = 2, = (7 - 2) = 5 and n = 19
Using the formula, Sn = n22a + n-1d, we get:
S19=1922×2+19-15                                 [ a = 2, d = 5 and n= 19]       = 1924+90=19×47=893
Hence, the sum of the first 19 terms of the given AP is 893.

Page No 500:

Question 2:

Find the sum of first 26 terms of the AP 1, 3, 5, 7, ..... .

Answer:

Here, a = 1, = (3-1) = 2 and n = 26
Using the formula, Sn = n22a + n-1d, we get:
S26=2621×2+26-12                                 [ a = 1, d = 2 and n= 26]       =13×2+50=13×52=676

Hence, the sum of the first 26 terms of the given AP is 676.

Page No 500:

Question 3:

Find the sum of first 18 terms of the AP 9, 7, 5, 3, ..... .

Answer:

Here, a = 9, = (7 - 9) = -2 and n = 18
Using the formula, Sn = n22a + n-1d, we get:
S18 = 1822×9+18-1×(-2)                                 [ a = 9, d = -2 and n= 18]       = 9×18-34=9×(-16)=-144

Hence, the sum of the first 18 terms of the given AP is -144.

Page No 500:

Question 4:

Find the sum (5 + 13 + 21 + ... + 181).

Answer:

Here, a = 5, d = (13 - 5) = 8 and l = 181
Let the total number of terms be n.
Then Tn = 181 
a + (n - 1)d = 181
              ⇒ 5 + (n - 1​) ⨯​ 8 = 181 
              ⇒ 8n = 184
              ⇒ n = 23
Hence, there are 23 terms in the AP.
∴ Required sum = n2a+l

                   = 2325 + 181 = 23×93 = 2139
Hence, the required sum is 2139.

Page No 500:

Question 5:

Which term of the AP 5, 9, 13, 17, ... is 81? Also find the sum (5 + 9 + 13 + 17 + ... + 81).

Answer:

In the given AP, first term, a = 5 and common difference, d = (9 - 5) = 4.
Let its nth term be 81.
Then Tn = 81
a + (n - 1)d = 81
              ⇒ 5 + (n - 1​) ⨯​ 4 = 81 
              ⇒ 4n = 80
              ⇒ n = 20
Hence, the 20th term of the given AP is 81.

∴ Required sum = n2a+l
                          = 2025 + 81 = 10×86 = 860
Hence, the required sum is 860.

Page No 500:

Question 6:

Which term of the AP 15, 11, 7, ... is (−13)? Also find the sum (15 + 11 + 7 + ... + (−13).

Answer:

In the given AP, first term, a  = 15 and common difference, d = (11 - 15) = -4.
Let its nth term be -13.
Then Tn = -13 
a + (n - 1)d = -13
              ⇒ 15 + ( n - 1​) ⨯​ (-4) = -13
              ⇒ 4n32
              ⇒ n = 8
Hence, the 8th term of the given AP is -13.
∴ Required sum = n2a+l
                       = 8215 + (-13) = 8
Hence, the required sum is 8.

Page No 500:

Question 7:

Find the sum of al two-digit whole numbers divisible by 3.

Answer:

All the two-digit whole numbers divisible by 3 are 12, 15, 18, 21,..., 99.
This is an AP in which a = 12, d = (15 - 12) = 3 and l = 99.
Let the number of terms be n.
Then first term (a) = 12 and common difference (d) = (15 - 12) = 3
Now, Tn = 99 
a + (n - 1)d = 99
              ⇒ 12 + ( n - 1​) ⨯​ 3 = 99
              ⇒ 3n = 90
              ⇒ n = 30

∴ Required sum = n2a+l
                         = 30212 + 99 = 15×111 = 1665
Hence, the required sum is 1665.

Page No 500:

Question 8:

Find the sum of all even numbers between 5 and 100.

Answer:

All the even numbers between 5 and 100 are 6, 8, 10, 12,..., 98.

This is an AP in which a = 6, d = (8 - 6) = 2 and l = 98.
Let the number of terms be n.
Then Tn = 98 
a + (n - 1)d = 98
              ⇒ 6 + ( n - 1​) ⨯​ 2 = 98
              ⇒ 2n94
              ⇒ n = 47

∴ Required sum = n2a+l
                        = 4726 + 98 = 47×52 = 2444
Hence, the required sum is 2444.

Page No 500:

Question 9:

Find the sum of all natural numbers between 100 and 500 which are divisible by 7.

Answer:

All the natural numbers less than 100 and divisible by 6 are 6, 12, 18, 24,...,96.

This is an AP in which a = 6, d = (12 - 6) = 6 and l = 96.
Let the number of terms be n.
Then Tn = 96 
a + (n - 1)d = 96
              ⇒ 6 + (n - 1​) ⨯​ 6 = 96
              ⇒ 6n = 96
              ⇒ n = 16

∴ Required sum = n2a+l
                         = 1626 + 96 = 8×102 = 816  
Hence, the required sum is 816.

Page No 500:

Question 10:

Find the sum of all natural numbers between 100 and 500 which are divisible by 7.

Answer:

All the natural numbers between 100 and 500 which are divisible by 7 are 105, 112, 117, 126,..., 497.

This is an AP in which a = 105, d = (112 - 105) = 7 and l = 497.
Let the number of terms be n.
Then Tn = 497 
a + (n - 1)d = 497
              ⇒ 105 + (n - 1​) ⨯​ 7 = 497
              ⇒ 7n = 399
              ⇒ n = 57

∴ Required sum = n2a+l
                         = 572105 + 497 = 57×301 = 17157
Hence, the required sum is 17157

Page No 500:

Question 11:

Find the sum of all multiples of 9 lying between 300 and 700.

Answer:

All the multiples of 9 lying between 300 and 700 are 306, 315, 324, 333, ..., 693.

This is an AP in which a = 306, d = (315 - 306) = 9 and l = 693.
Let the number of terms be n.
Then Tn = 693
a + (n - 1)d = 693
⇒ 306 + (n - 1​) ⨯​ 9 = 693
⇒ 9n = 396
⇒ n = 44

∴ Required sum = n2a+l
                   = 442306 + 693 = 22×999 = 21978  
Hence, the required sum is 21978.

Page No 500:

Question 12:

Find the sum of all three-digit natural numbers which are divisible by 13.

Answer:

All three-digit numbers which are divisible by 13 are 104, 117, 130, 143,..., 988.
 
This is an AP in which a = 104, d = (117 - 104) = 13 and l = 988
Let the number of terms be n.
Then Tn = 988 
a + (n - 1)d = 988
⇒ 104 + (n -1​) ⨯​ 13 = 988
⇒ 13n = 897
⇒ n = 69

∴ Required sum = n2a+l
                          = 692104 + 988 = 69×546 = 37674  
Hence, the required sum is 37674.

Page No 500:

Question 13:

Find the sum of first 15 multiples of 8.

Answer:

The first 15 multiples of 8 are 8, 16, 24, 32,...
This is an AP in which a = 8, d = (16 - 8) = 8 and n = 15.
Thus, we have:
l = a+n-1d  = 8+(15-1)8  = 120
 
∴ Required sum = n2a+l

                   = 1528 + 120 = 15×64 = 960  
Hence, the required sum is 960.

Page No 500:

Question 14:

Find the sum of all odd numbers between 0 and 50.

Answer:

All odd numbers between 0 and 50 are 1, 3, 5, 7, ..., 49.
This is an AP in which a = 1, d = (3 - 1) = 2 and l = 49.
Let the number of terms be n.
Then, Tn = 49 
a + (n - 1)d = 49
              ⇒ 1 + (n - 1​) ⨯ 2 = 49
              ⇒ 2n50
              ⇒ n = 25

∴ Required sum = n2a+l

                   = 2521 + 49 = 25×25 = 625  
Hence, the required sum is 625.

Page No 500:

Question 15:

Find the sum of first hundred even natural numbers which are divisible by 5.

Answer:

The first few even natural numbers which are divisible by 5 are 10, 20, 30, 40, ...
This is an AP in which a = 10, d = (20 - 10) = 10 and n = 100
The sum of n terms of an AP is given by
Sn=n22a+n-1d    =1002×[ 2×10+100-1×10 ]              a = 10, d = 10 and n = 100    =50×20+990=50×1010=50500

Hence, the sum of the first hundred even natural numbers which are divisible by 5 is 50500.

Page No 500:

Question 16:

Find the sum of all 3-digit natural numbers which are divisible by 13.

Answer:

All three-digit numbers which are divisible by 13 are 104, 117, 130, 143, ..., 988.
This is an AP in which a = 104, d = (117 - 104) = 13 and l = 988.
Let the number of terms be n.
Then
Tn = 988
 ⇒
a + (n - 1)d = 988
              ⇒ 104 + (n - 1​) ⨯​13 = 988
              ⇒ 13n = 897
              ⇒ n = 69

∴ Required sum =  n2a+l
                     =692104 + 988 = 69 ×546 = 37674
  Hence, the required sum is 37674.
 

Page No 500:

Question 17:

Find the sum of 51 terms of the AP whose second term is 2 and the 4th term is 8.

Answer:

Let a be the first term and d be the common difference of the given AP.
Then T2 = 2 and T4 = 8
 ⇒ a + (2 - 1)d = 2 and a + (4  - 1)d = 8
 i.e., a + d = 2          ...(i)
also, a + 3d = 8      ...(ii)
On subtracting (i) from (ii), we get:
2d = 6  
d = 3
Putting d = 3 in (i), we get:
a = -1
a = -1, d = 3 and n = 51

Now, sum of the first 51 terms = n2 [ 2a + n-1d]
                              =512×2×-1+51-1×3=512×-2+150=512×148=51×74=3774    
            
  Hence, the required sum is 3774.

Page No 500:

Question 18:

If the 5th and 12th terms of an AP −4 and −18 respectively, find the sum of first 20 terms of the AP.

Answer:

Let a be the first term and d be the common difference of the given AP.
Then T5 = -4 and T12 = -18
 a + (5 - 1) d = -4 and a + (12 - 1) d = -18
i.e., a + 4d = - 4     ...(i)
also, a +11d = -18          ...(ii)
On subtracting (i) from (ii), we get:
7d = -14
⇒ d = -2
Putting d = -2 in (i), we get:
a = 4
∴ a = 4, d = -2 and n = 20
Sum of the first 20 terms  = n2 [ 2a + n-1d]
                              =202×2×4+20-1×-2=  10×8-38=10×-30=-300    
            
  Hence, the required sum is -300.

Page No 500:

Question 19:

How many terms of the AP 21, 18, 15, ... must be added to get the sum 0?

Answer:

Here, a = 21 and d = (18 - 21) = -3
Let the required number of terms be n.
Then, Sn  = 0
⇒  n2 [ 2a + n-1d] = 0
         n2×2×21 + n-1×-3= 0   n2×45 -3n = 0  45n - 3n2 = 0  3n = 45n = 15    
            
  Hence, the required number of terms is 15.

Page No 500:

Question 20:

Find the number of terms of the AP 63, 60, 57, ... so that their sum is 693. Explain the double answer.

Answer:

Here, a = 63 and d = (60 - 63) = -3
Let the required number of terms be n.
Then, Sn  = 693
n2 [ 2a + n-1d] = 693
         n2×2×63 + n-1×-3= 693   n2×129 -3n = 693   3n2 - 129n + 1386= 0  n2  - 43n + 462 = 0n2  - 21n - 22n + 462 = 0n(n -21) -22(n -21) = 0(n-21) (n-22) = 0n = 21 or n = 22    
 
  ∴ Sum of the first 21 terms = sum of the first 22 terms = 693
 This means that the term T22 is 0.

Page No 500:

Question 21:

Find the number of terms of the AP 64, 60, 56, ... so that their sum is 544. Explain the double answer.

Answer:

Here, a = 64 and  d = (60 - 64) = -4
Let the required number of terms be n.
Then, Sn = 544 
n2 [ 2a + n-1d] = 544
         n2×2×64 + n-1×-4= 544   n2×132 -4n = 544   2n2 - 66n + 544= 0  n2  - 33n + 272 = 0n2  - 17n - 16n + 272 = 0n(n -17) -16(n -17) = 0(n-17) (n-16) = 0n = 17 or n = 16    
 
  ∴ Sum of the first 16 terms = sum of the first 17 terms = 544
 This means that the term T17 is 0.

Page No 500:

Question 22:

Find the number of terms of the AP 18, 15, 12, ... so that their sum is 45. Explain the double answer.

Answer:

Here, a  = 18 and  d = (15- 18) = -3
Let the required number of terms be n.
Then, Sn  = 45 
⇒  n2 [ 2a + n-1d] = 45
        n2×2×18 + n-1×-3=45n2×39 -3n = 453n2-39n+90=0n2-13n+30=0n2-10n-3n+30=0n(n-10)-3(n-10)=0(n-10)(n-3)=0n=3 or n=10    
 
  ∴ Sum of the first 3 terms = sum of the first 10 terms = 45
 This means that the sum of all terms from 4th to 10th is zero.



Page No 501:

Question 23:

If the nth term of an AP is (4n + 1), find the sum or the first 15 terms of this AP. Also find the sum of its n terms.

Answer:

Given: Tn = (4n + 1).
 Now, T1 = (4 ⨯ 1 + 1) = 5
 T2 = (
4 ⨯ 2 + 1) = 9
 T15 = (
4 ⨯ 15 + 1) = 61
Hence, a = 5, d = (9 - 5) = 4 and l = 61

 
∴ S15 = n2a+l
      = 152×5+61 =15×33 = 495

Hence, the sum of the first 15 terms is 495.
The sum of the first n terms is given by
 Sn=n22a+n-1d    =n22×5+n-1×4=n26+4n=3n+2n2

Page No 501:

Question 24:

The sum of the first n terms of an AP is given by Sn = (2n2 + 5n). Find the nth term of the AP.

Answer:

Given: Sn = (2n2 + 5n)              ...(i)
Replacing n by (n - 1) in (i), we get:
Sn-1 = 2(n - 1)2 + 5(n - 1)
        = 2(n2 - 2n + 1) + 5n - 5
        = 2n2n - 3
∴​ Tn = (Sn - Sn-1)
​        = 
(2n2 + 5n) - (2n2 + n - 3 ) =  4n +3
∴ nth term, Tn = (
4n +3)

 

Page No 501:

Question 25:

If the sum of the first n terms of an AP is given by Sn = (3n2 − n), find its (i) nth term, (ii) first term, and (iii) common difference.

Answer:

Given: Sn = (3n2n)               ...(i)
Replacing n by (n - 1) in (i), we get:
Sn-1 = 3(n - 1)2 - (n - 1)
        = 3(n2 - 2n + 1) - + 1
        = 3n2 - 7n + 4
(i) Now,​ Tn = ( Sn - Sn-1)
​    = 
(3n2 - n) - (3n2 - 7n + 4 ) = 6n - 4
∴ nth term, Tn = (
6n - 4)                ...(ii)

(ii) Putting n = 1 in (ii), we get:
T1= (6 ⨯ 1) - 4 = 2
 

(iii) Putting n = 2 in (ii), we get:
T2= (6 ⨯ 2) - 4 = 8
∴ Common difference, d = T2 - T1 = 8 - 2 = 6

Page No 501:

Question 26:

The sum of n terms of an AP is 49 and that of first 17 terms is 289, find the sum of first n terms.

Answer:

 Sn = 5n22+3n2 = 125n2 + 3n        ...(i)Replacing n by (n-1) in (i), we get:Sn-1=12×5n-12+3n-1         =12×5n2-10n+5+3n-3]=12×[5n2 -7n+2]Tn=Sn-Sn-1        =125n2+3n-12×[5n2 -7n+2]       =1210n -2=5n-1               ...(ii)


Putting n = 20 in (ii), we get:
T20 = (5 ⨯ 20) - 1 = 99
Hence, the 20th term is 99.

Page No 501:

Question 27:

If the sum of first 7 terms of an AP is 49 and that of first 17 terms is 289, find the sum of first n terms.

Answer:

Let a be the first term and d be the common difference of the given AP.
Then we have:
Sn =  n22a + n-1dS7 = 722a + 6d  = 7[a +3d]  S17 =1722a + 16d  = 17[a +8d]  

However, S7 = 49 and S17 = 289
Now, 7[a + 3d] = 49 
a + 3d = 7           ...(i)
Also, 17[a + 8d] = 289  
​⇒ a + 8d = 17           ...(ii)

Subtracting (i) from (ii), we get:

5d = 10
d = 2

Putting d = 2 in (i), we get:
a + 6 = 7 
a = 1
Thus, a = 1 and d = 2

∴ Sum of n terms of AP = n22×1 + n-1×2 = n [ 1+ (n-1)] = n2

Page No 501:

Question 28:

The sum of the first 9 terms of an AP is 81 and the sum of its first 20 terms is 400. Find the first term and the common difference of the AP.

Answer:

Let a be the first term and d be the common difference of the given AP.
Then we have:
Sn =  n22a + n-1dS9 = 922a + 8d  = 9[a +4d]  S20 =2022a + 19d  = 10[2a +19d]  

However, S9 = 81 and S20 = 400

Now, 9[a + 4d] = 81
⇒ a + 4d = 9                 ...(i)
Also, 10[ 2a +19d ] = 400
⇒2a + 19d = 40           ...(ii)

Multiplying (i) by 2 and subtracting the result from (ii), we get:

11d = 22
 ⇒ d = 2

Putting d = 2 in (i), we get:
a + 8 = 9 ⇒a = 1
Thus, a = 1 and d = 2
 

Page No 501:

Question 29:

The first and last terms of an AP are 4 and 81 respectively. If the common difference is 7, how many terms are there in the AP and what is their sum?

Answer:

Here, a = 4, d = 7 and l = 81
Let the nth term be 81.
Then Tn = 81
⇒ a + (n - 1)d = 4 + (n - 1)7 =  81
⇒ (n - 1)7 = 77
⇒ (n - 1) = 11
n = 12
Thus, there are 12 terms in the AP.

The sum of n terms of an AP is given by
Sn =  n2a + lS12 = 1224 + 81 = 6×85=510 


Thus, the required sum is 510.

Page No 501:

Question 30:

In an AP the first term is 22, nth term is −11 and sum to first nth terms is 66. Find n and d, the common difference.

Answer:

Here, a = 22, Tn = -11 and Sn = 66
Let d be the common difference of the given AP. 
Then Tn = -11 
⇒ a + (n - 1)d = 22 + (n - 1)d = -11
⇒ (n - 1)d = -33        ...(i)

The sum of n terms of an AP is given by
Sn = n22a + n-1d = 66        n22×22 + (-33)  = n2 ×11= 66        n = 12        [Substituting the value of (n - 1)d from (i)]
Putting the value of n in (i), we get:
11d = -33
d = -3
Thus, n = 12 and d = -3

Page No 501:

Question 31:

In an AP the first term is 2, the last term is 29 and sum of the term is 155. Find the common difference of the AP.

Answer:

Here, a = 2, l = 29 and Sn = 155
Let d be the common difference of the given AP and n be the total number of terms. 
Then Tn = 29
a + (n - 1)d = 29  
⇒ 2 + (n - 1)d = 29​                ...(i)

The sum of terms of an AP is given by
Sn =  n2a + l = 155        n22 + 29  = n2 ×31= 155        n = 10        
Putting the value of n in (i), we get:
⇒ 2 + 9d = 29
⇒ 9d = 27
d = 3
Thus, the common difference of the given AP is 3.

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Question 32:

Find the sum of the following:
1-1n+1-2n+1-3n+...upto n terms.

Answer:

On simplifying the given series, we get:

1-1n+1-2n+1-3n+...n terms=(1+1+1+...n terms)-1n+2n+3n+...+nn=n-1n+2n+3n+...+nnHere, 1n+2n+3n+...+nn is an AP whose first term is 1n and the common difference is 2n-1n=1n.The sum of n terms of an AP is given bySn=n22a+n-1d      =n-n22×1n+(n-1)×1n                             =n-n22n+n-1n=n-n2n+1n       =n-n+12=n-12

Page No 501:

Question 33:

The production of TV in a factory increase uniformly by a fixed number every year. It produced 8000 acts in 6th year and 11300 in 9th year. Find the production in (i) first year (ii) 8th year and  (iii) 6th year.

Answer:

(i) As the production of TV sets increases uniformly by a fixed number every year, the number of TV sets produced in the 1st year, 2nd year, 3rd year, ... will form an AP. Let the first term be a, common difference of the AP be d and the number of TV sets produced in the nth year be Tn.

Then T6 = 8000    
⇒ a + 5d  =  8000          ...(i)
Also, T9 = 11300
a + 8d = 11300          ...(ii)

On subtracting (i) from (ii), we get:
⇒ 3d = 3300
⇒​ d = 1100

Putting the value of d in (i), we get:
⇒ a + 5 ⨯ 1100 = 8000
a = 8000 - 5500 = 2500

Therefore, the production of TV sets in 1st year is 2500.

(ii) Now, production of TV sets in the 8th year is given by
    T8a + 7d = 2500 + 7 ⨯ 1100 = 10200

(iii)  Sum of terms of an AP is given by
       Sn=n2a+l S6=622500 + 8000=3×10500=31500            
Thus, the total production of TV sets in the 6th years is 31500.



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Question 34:

A sum of Rs 2800 is to be used to award four prizes. If each prize after the first is Rs 200 less than the preceding prize, find the value of each of the prizes.

Answer:


Let the amounts of the four prizes be Rs a, Rs (a - 200), Rs ( a - 400) and Rs ( a - 600).
Then, a + a - 200 +  - 400 + a  - 600 = Rs 2800
⇒ 4a - 1200 = 2800
⇒ 4a = 4000
 ⇒ a = Rs 1000
Putting the value of a in each term, we get the amounts of the four prizes as follows:​
Rs 1000, Rs (1000 - 200) = Rs 800, Rs (1000 - 400) = Rs 600 and Rs (1000 - 600) = Rs 400

Thus, the amounts of prizes are Rs 1000, Rs 800, Rs 600 and Rs 400.

Page No 502:

Question 35:

200 logs are stacked in such a way that there are 20 logs in the bottom row, 19 in the next row, 18 in the next row, and so on. In how many rows 200 logs are placed and how many logs are there in the top row?

Answer:

From the bottom, the number of logs in adjacent rows are 20, 19, 18,...
Since the logs decreases uniformly by 1 in each row, the number of logs in the rows form an AP.
Here, a = 20, d = 19 - 20 = -1 
Let there be n rows such that Sn = 200.
The sum of n terms of an AP is given by
 Sn = n22a + n-1dHere, Sn = 200Sn =200  n2×2×20 + n-1×-1 = 200 n2×[ 40 - n +1] =200n×[41 - n] = 400n2 - 41n +400 =  0 n2 - 25n - 16 n+400 =0n(n - 25) - 16 (n - 25)=0 n = 25 or n = 16

i.e., the number of rows is either 16 or 25.
For n = 25, we have:
T25 = a + 24 d = 20 + 24 ⨯ (-1) = -4, which is not possible.
Therefore, n = 25 is not acceptable.

For n = 16, we have:
T16 = a +15d = 20 + 15 ⨯ (-1)​ = 5
Thus, there are 16 rows and 5 logs are placed in the top row.



Page No 503:

Question 1:

If 45,a, 2 are three consecutive terms of an AP, find the value of a.

Answer:

If  45, a and 2 are three consecutive terms of an AP, then we have:
a45 = 2 - a
⇒ 2a  = 2 + 45
2a145
a = 75

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Question 2:

If (x+2),2x,(2x+3) are three consecutive terms of an AP, find the value of x.

Answer:

If (x+2),2x and (2x+3) are three consecutive terms of an AP, then we have: 
2x - (x+2) = (2x+3) - 2x
⇒ x - 2 = 3
x = 5​
Hence, the value of x is 5.

Page No 503:

Question 3:

For what value of p are (2p+1),13,(5p-3) three consecutive terms of an AP?

Answer:

Let (2p+1), 13,(5p-3) be three consecutive terms of an AP.
Then
13(2p+1) =(5p - 3) - 13
⇒ 7p = 28
p = 4​
∴ When p = 4,  
(2p+1), 13 and(5p-3) form three consecutive terms of an AP.

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Question 4:

For what value of p are (2p-1), 7 and 3p three consecutive terms of an AP?

Answer:

Let (2p-1),7 and 3p be three consecutive terms of an AP.
Then
7 - (2p - 1) = 3p - 7
5p = 15
p = 3​
∴ When p = 3,
(2p -1), 7 and 3p form three consecutive terms of an AP.

Page No 503:

Question 5:

The nth term of an AP is (3n + 5). Find its common difference.

Answer:

We have:
Tn = (3n + 5)
Common difference = T2 - T1
​T1 = 3 ⨯ 1 + 5 = 8
T2 = ​3 ⨯ 2 + 5 = 11
d = 11 - 8 = 3
Hence, the common difference is 3.

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Question 6:

The nth term of an AP is (7 − 4n). Find its common difference.

Answer:

We have:
Tn = (7 - 4n)
Common difference = T2 - T1
​T= 7 - 4 ⨯ 1 = 3
T2 = ​7 - 4 ⨯ 2 = -1
d = -1 - 3 = -4
Hence, the common difference is -4.

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Question 7:

The first term of an AP is p and its common difference is q. Find its 10th term.

Answer:

Here, a = p and d = q
Now, Tn = a + (n - 1)d  
Tn = p + (n - 1)q
∴ ​T10p + 9q

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Question 8:

Write the next term of the AP 8,18,32,..... .

Answer:

The given AP is 8, 18, 32, ... On simplifying the terms, we get: 22, 32, 42, ...Here, a=22 and d=(32-22)=2 Next term, T4=a+3d=22+32=52=50

Page No 503:

Question 9:

Write the next term of the AP 2,8,18,..... .

Answer:

:The given AP is 2, 8, 18,... On simplifying the terms, we get: 2, 22, 32, ...Here, a=2 and d=(22-2)=2 Next term, T4=a+3d=2+32=42=32

Page No 503:

Question 10:

Which term of the AP 21, 18, 15,... is zero?

Answer:

In the given AP, first term, a = 21 and common difference, d = (18 - 21) = -3
Let's its nth term be 0.
Then
Tn = 0
 ⇒ 
a + (n - 1)d = 0
              ​ ⇒ 21 + (n - 1) ⨯ (-3) = 0
               ⇒ 24 - 3n = 0
               ⇒ 3n = 24
               ⇒ n = 8
Hence, the 8th term of the given AP is 0.
 

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Question 11:

Which term of the AP 25, 20, 15,... is the first negative term?

Answer:

Here, a  = 25 and d = (20 - 25) = -5
Let the nth term of the given AP be the first negative term.
Then
Tn < 0
 ⇒ 
a + (n - 1)d < 0
              ​ ⇒ 25 + (n - 1) ⨯ (- 5) < 0
               ⇒ 30 - 5n < 0
               ⇒  30 < 5n

               ⇒ n > 6
               ∴ n = 7

Hence, the 7th term of the given AP is the first negative term.
 

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Question 12:

Find the 6th term from the end of the AP 5, 7, 9, ...., 201.

Answer:

Here, a = 5, d = (7 - 5) = 2, l = 201 and n = 6th
Now, nth term from the end = [l - (n - 1)d]  
6th term from the end = [201 - (6 - 1) ⨯ 2]
                                 = [201 - (5 ⨯ 2)] = (201 - 10) = 191
Hence, the 6th term from the end of the AP is 191.

Page No 503:

Question 13:

Find the sum of first n natural numbers.

Answer:

The first n natural numbers are 1, 2, 3, 4, 5, ..., n.

Here, a = 1 and d = (2 - 1) = 1
 
Sum of n terms of an AP is given bySn=n22a+n-1d      = n2×[2×1 +n-1×1]     = n2×[2+n -1] = n2×n+1 = nn+12

Page No 503:

Question 14:

Find the sum of first n even natural numbers.

Answer:

The first n even natural numbers are 2 ,4, 6, 8, 10, ..., n.

Here, = 2 and d = (4 - 2) = 2
 
Sum of n terms of an AP is given bySn = n22a + n-1d      = n2×[2×2 +  n-1×2]     = n2×[4+ 2n -2] = n2×2n+2 =  nn+1
Hence, the required sum is n(n+1).

Page No 503:

Question 15:

Find the sum of the first n odd natural numbers.

Answer:

The first n odd natural numbers are 1, 3, 5, 7, 9, ..., n.
Here, = 1 and d = (3 - 1) = 2
 
Sum of n terms of an AP is given bySn = n22a + n-1d      = n2×[2×1 +  n-1×2]     = n2×[2+ 2n -2] = n2×2n =  n22
Hence, the required sum is n2.



Page No 504:

Question 1:

If (x+2),2x,(2x+3) are three consecutive terms of an AP, then x = ?
(a) 3
(b) 4
(c) 5
(d) 6

Answer:

(c) 5

If (x+2),2x and (2x+3) are three consecutive terms of an AP, then we have: 
2x - (x+2) = (2x+3) - 2x
⇒ x - 2 = 3
x = 5​

Page No 504:

Question 2:

If (2p+1),13,(5p-3) are three consecutive terms of an AP, then p = ?
(a) 4
(b) 5
(c) 6
(d) 3

Answer:

(a) 4
If (2p + 1), 13 and(5p - 3) are three consecutive terms of an AP, then we have: 
13(2p+1) = (5p-3) - 13
122p = 5p - 16
7​p = 28
⇒ p = 4

Page No 504:

Question 3:

If (k+1), 3k, (4k+2) are three consecutive terms of an AP, then k = ?
(a) 0
(b) 1
(c) 2
(d) 3

Answer:

(d) 3
If (k+1),3k and(4k+2) are three consecutive terms of an AP, then we have:
3k - (k + 1) = (4k + 2) - 3k
⇒ 2k - 1 = k +2
k = 3


Page No 504:

Question 4:

If 45,a,2 are three consecutive terms of an AP, then a = ?
(a) 5
(b) 54
(c) 95
(d) 75

Answer:

(d) 75

If 45a and 2 are three consecutive terms of an AP, then we have:
a - 45 = 2 - a
⇒ 2a2 + 45 
2a = 145 
a = 75

Page No 504:

Question 5:

What is the next term of the AP 2,8,18,...?
(a) 24
(b) 28
(c) 30
(d) 32

Answer:

(d)   3232
The given AP is 2, 8, 18, ... On simplifying the terms, we get: 2, 22, 32, ...Here, a=2 and d=(22-2)=2 Next term, T4=a+3d=2+32=42=32

Page No 504:

Question 6:

What is the next term of the AP 8,18,32,...?
(a) 40
(b) 48
(c) 50
(d) 54

Answer:

(c) 50

The given AP is 8, 18, 32, ... On simplifying the terms, we get:22, 32, 42, ...Here, a=22 and d=(32-22)=2 Next term, T4=a+3d=22+32=52=50

Page No 504:

Question 7:

The first term of an AP is p and its common difference is q. What is its 10th term?
(a) p + 10q
(b) q + 10p
(c) p + 9q
(d) q + 9p

Answer:

(c) p + 9q

Here, a = p and d = q
Now, Tn = a + (n - 1)d  
 Tn = p + (n - 1)q
⇒ ​T10 = p + 9q

Page No 504:

Question 8:

The number −11, −7, −3, 1, 5, ... are
(a) in AP with d = −18
(b) in AP with d = −14
(c) in AP with d = −4
(d) not in AP

Answer:

(c) in AP with d = 4

In the given progression, [-7 - (-11)] = [-3 - (-7)] = [1 -(-3)] = 4 (constant)
Thus, each term differs from its preceding term by 4. 
So, the given progression is an AP with = 4. 

Page No 504:

Question 9:

The 30th term of the AP 10, 7, 4, .... is
(a) −87
(b) 87
(c) 77
(d) −77

Answer:

(d) - 77

   The given AP is 10, 7, 4, ...
    Here, first term, a = 10 and common difference, d = (7 - 10) = -3
    Now, T30 a + (30 - 1)d =  a + 29d

         = 10 + 29 ⨯ (-3) = -77
   Hence, 30th term = -77

Page No 504:

Question 10:

The 11th term of the AP −3, -12,2,.... is
(a) 22
(b) 28
(c) −38
(d) 4812

Answer:


(a) 22
   The given AP is -3, -12, 2, ...
    Here, first term, a = -3 and common difference, d-12 - -3 = 3 - 12 = 212
    Now, T11=a+11-1d=a+10d       =-3+10×212=-3+10×52=-3+25=22


   ∴ 11th term  =  22

Page No 504:

Question 11:

The first term of an AP is 6 and its common difference is −3. What is the 16th term of the AP?
(a) 51
(b) 39
(c) −39
(d) −42
 

Answer:

(c) -39

Here, a = 6 and d = -3
Then, T16 = a + 15d = 6 + 15 ⨯ (-3) = -39
Hence, the 16th term of given AP is -39.

Page No 504:

Question 12:

The first two terms of an AP are −5 and 6. Its 21st term is
(a) 115
(b) −115
(c) −215
(d) 215

Answer:

(d) 215
Here, T1 or a = -5 and T2 = 6
Then, d = T2​ - T1 = 6 - (-5) = 11
Now, T21 = a + 20d = -5 + 20 ⨯ 11 = 215
Hence, the 21st term of the given AP is 215.

Page No 504:

Question 13:

Which term of the AP 21, 18, 15,...is −81?
(a) 25th
(b) 18th
(c) 35th
(d) 46th

Answer:

(c) 35th
In the given AP, we have 
a  = 21 and d =  (18 - 21) =  -3
Let the nth term be -81.
Then
Tn = -81
 ⇒ 
a + (n - 1)d = -81
              ​ ⇒ 21 + (n - 1) ⨯ (-3) = -81
               ⇒ 24 - 3n = -81
              ⇒ 3n = 105
              ⇒ n = 35
Hence, the 35th terms is -81.



Page No 505:

Question 14:

Which term of the AP 72, 63, 54, ... is 0?
(a) 8th
(b) 9th
(c) 10th
(d) 11th

Answer:

(b) 9th

In the given AP, we have 
a  = 72 and d = (63 - 72) =  - 9
Let nth terms be 0.
Then
Tn = 0
⇒ 
a + (n -1) d = 0
⇒ 72 + (n - 1) ⨯(-9) = 0
⇒ 81 - 9n = 0
⇒ 9n =  81
n = 9
Hence, the 9th term is 0.

Page No 505:

Question 15:

Which term of the AP 25, 20, 15, ... is the first negative term?
(a) 9th
(b) 8th
(c) 7th
(d) 10th

Answer:

(c) 7th

Here, 
a  = 25 and d =  (20 - 25) = - 5
Let the nth term of the given AP be the first negative term.
Then
Tn < 0
⇒ a + (n -1) d < 0
⇒ 25 + (n - 1) ⨯ (-5) < 0
⇒ 30 - 5n < 0
⇒ 30 <  5n
5n > 30
n > 6
∴ n = 7

Hence, the 7th term is the first negative term.

Page No 505:

Question 16:

If the nth term of an AP is (3n + 5), then its common difference is
(a) 5
(b) 4
(c) 3
(d) 2

Answer:

(c) 3
Here, Tn = 3n + 5
Then, T1 = 3 ⨯ ​1 + 5 = 8
Also, T2 = 3 ⨯ ​2 + 5 = 11​
∴ d = T2 - T1 = 11 - 8 = 3

Hence, the common difference is 3.

Page No 505:

Question 17:

If the nth term of an AP is (7 − 4n), then its common difference is
(a) 4
(b) −4
(c) 3
(d) −3

Answer:

(b) -4
Here, Tn = 7 - 4n  
Then T1 = 7 - 4 ⨯ ​1 = 3
Also, T2 = 7 - 4 ⨯ ​2 = -1​
d = T2 - T1 = -1 - 3 = -4

Hence, the common difference is -4.

Page No 505:

Question 18:

What is the common difference of an AP in which T18 −T14 = 32?
(a) 8
(b) −8
(c) 4
(d) −4

Answer:

(a) 8

The common difference of two terms is constant in an AP.
So, we have:
T18 - T17 = d       ...(i)
T17 - T16 = d       ...(ii)
T16 - T15 = d       ...(iii)
T15 - T14 = d       ...(iv)

On adding the four equations, we get:
T18 - T14 = 4d
However, ​T18 - T17 = 32
i.e., 4d = 32
⇒​ d = 8

Page No 505:

Question 19:

In an AP, the 7th term is 4 and the common difference is −4. What is its first term?
(a) 16
(b) 20
(c) 24
(d) 28

Answer:

(d) 28

T7 = a + 6d 
a + 6 ⨯ (-4) = 4
a = 4 + 24 = 28
Hence, the first term is 28.

Page No 505:

Question 20:

The 4th term of an AP is 14 and its 12th term is 70. What is its first term?
(a) 7
(b) 10
(c) −7
(d) −10

Answer:

( c) -7
T4 = a + 3d = 14       ...(i)
T12a + 11d = 70     ...(ii)
On subtracting (i) from (ii), we get:
8d = 56
d = 7

On putting the value of d in (i), we get:
a + 3 × 7 = 14
⇒ a = -7
Hence, the first term is -7.

Page No 505:

Question 21:

The 8th term of an AP is 17 and its 14th term is 29. The common difference of the AP is
(a) 3
(b) 2
(c) 5
(d) −2

Answer:

( b) 2
T8 = a + 7d = 17     ...(i)
T14 = a + 13d = 29    ...(ii)

On subtracting (i) from (ii), we get:
⇒ 6d = 12
d = 2
Hence, the common difference is 2.

Page No 505:

Question 22:

The 17th term of an AP exceeds its 10th term by 21. The common difference of the AP is
(a) 3
(b) 2
(c) −3
(d) −2

Answer:

( a) 3
T10 = a + 9d   
T17 = a + 16d 

Also, a + 16d = 21 + T10
     ⇒ a + 16d = 21 + a + 9d
     ⇒ 7d = 21
     ⇒ d = 3
Hence, the common difference of the AP is 3.

Page No 505:

Question 23:

If the 3rd and 9th terms of an AP are 4 and −8 respectively, then which term of the AP is 0?
(a) 5th
(b) 7th
(c) 11th
(d) 13th

Answer:


( a) 5th
T3 = a + 2d = 4            ...(i)
 T9 = a + 8d = -8          ...(ii)
 On subtracting (i) from (ii), we get:
⇒ 6d = -12
d = -2
Putting the value of d in (i), we get:
a - 4 = 4
a = 8
Let the Tn term be 0.
Then Tn = 0
⇒ ​a + (n - 1)d = 0
⇒ 8 + (n - 1) ⨯ (-2) = 0
⇒ 2n = 10
n = 5
Hence, the 5th term of the AP is 0.

Page No 505:

Question 24:

If 7 times the 7th term of an AP is equal to 11 times the 11th term, then its 18th term will be
(a) 7
(b) 11
(c) 18
(d) 0

Answer:

( d) 0
In the given AP, let the first term be a and the common difference be d.
We have Tn = a + (n - 1)d 
​Then
T7 = a + (7 - 1)d  = a + 6d   

 Also, T11 = a + (11 - 1)d = a + 10d               
Now, (7 ⨯ T7) = (11 ⨯ T11
⇒ 7[ 
a + 6d] = 11[a + 10d]
⇒ 7a + 42d = 11a + 110 d
⇒ 4a = -68d
a = -17d

∴ T18 = a + (18 - 1)d =​ a + 17d = -17d + 17d = 0             [ ∵ a = - 17d]​
 ​Hence, the 18th term is 0 (zero).

Page No 505:

Question 25:

Which term of the AP 21, 42, 63, 84, ... is 210?
(a) 9th
(b) 10th
(c) 11th
(d) 12th

Answer:

(b) 10th

Here, a = 21 and d = (42 - 21) = 21
Let 210 be the nth term of the given AP.
Then ​Tn = 210 
⇒​ a + (n - 1)d = 210
⇒​ 21 + (n - 1) ⨯ 21= 210
⇒ 21n = 210
n = 10
Hence, 210 is the 10th term of the AP.

Page No 505:

Question 26:

If 4, x1, x2, x3, 28 are in AP, then x3 = ?
(a) 19
(b) 23
(c) 22
(d) cannot be determined

Answer:

(c) 22

Here, a = 4, l = 28 and n = 5
Then, ​T5 = 28
⇒​ a + (n - 1)d = 28
⇒​ 4 + (5 - 1)d= 28
⇒ 4d = 24
d = 6
Hence, x3 = 28 - 6 = 22

Page No 505:

Question 27:

The 20th term from the end of the AP 3, 8, 13, ..., 253 is
(a) 163
(b) 158
(c) 153
(d) 148

Answer:

(b) 158
Here, a = 3, d = (8 - 3) = 5, l  = 253 and n = 20th
Now, nth term from the end = [l - (n - 1)d]  
20th term from the end = [253 - (20 - 1) ⨯ 5]
                                 = [253 - (19 ⨯ 5)] = (253 - 95) = 158
Hence, the 20th term from the end of the AP is 158.



Page No 506:

Question 28:

The 4th term from the end of AP −11, −8, −5, ... ,49 is
(a) 58
(b) 43
(c) 40
(d) 37

Answer:

(c) 40

Here, a = -11 and d = [-8 - (-11)] = 3, l = 49 and n = 4th
Now, nth term from the end = [l - (n - 1)d]  
4th term from the end = [49 - (4 - 1) ⨯ 3]
                                 = [49 - (3 ⨯ 3)] = (49 - 9) = 40
Hence, the 4th term from the end of the AP is 40.

Page No 506:

Question 29:

The second term of an AP is 13 and its 5th term is 25. What is its 17th term.
(a) 73
(b) 77
(c) 69
(d) 81

Answer:

( a) 73

T2 = a + d  = 13              ...(i)
T5 = a + 4d  = 25            ...(ii)
On subtracting (i) from (ii), we get:
3d = 12
d = 4
On putting the value of d in (i), we get:
a + 4= 13
a = 9

Now, T17 = a +16d = 9 + 16 ⨯ 4 = 73
Hence, the 17th term is 73.

Page No 506:

Question 30:

The sum of first 16 terms of the AP 10, 6, 2, ..., is
(a) 320
(b) −320
(c) −352
(d) −400

Answer:

(b) - 320

Here, a = 10, = (6 - 10) = -4 and n = 16
Using the formula, Sn = n22a + n-1d, we get:
S16 = 1622×10 + 16-1 ×(-4)                              [ a = 10, d = -4 and n= 16]       = 8×20  - 60 = 8× -40=-320
Hence, the sum of the first 16 terms of the given AP is -320.

Page No 506:

Question 31:

The sum of first 20 terms of the AP 1, 3, 5, 7, 9, ... is
(a) 264
(b) 400
(c) 472
(d) 563

Answer:

(b) 400

Here, a = 1, = (3 - 1) = 2 and n = 20
Using the formula, Sn = n22a + n-1d, we get:
S20 = 2022×1 + 20-1 ×2                                 [ a = 1, d = 2 and n= 20]       = 10×2  + 38 = 10× 40=400
Hence, the sum of the first 20 terms of the given AP is 400.

Page No 506:

Question 32:

(5 + 13 + 21 + ... + 181) = ?
(a) 2476
(b) 2337
(c) 2219
(d) 2139

Answer:

(d) 2139
 
Here, = 5, d = (13-5) = 8 and l = 181
Let the number of terms be n.
Then Tn = 181 
        
a + (n-1d = 181
         ⇒ 5 + ( n -1​) ⨯​ 8 = 181
         ⇒  8n = 184
         ⇒ n = 23
 
∴ Required sum =   n2a+l
                     =2325 + 181 = 23 ×93 = 2139
  Hence, the required sum is 2139.

Page No 506:

Question 33:

The sum of n terms of the AP 2,8,18,32,....is
(a) 1
(b) 2n (n + 1)
(c) 12n(n+1)
(d) 12n(n+1)

Answer:

(d) 12nn+1

 Here, a = 2 and d = 8 -2 = 22 -2 =2 Sn = n22a + n-1d    n22×2 + n-1×2      n222 + n -1     n2 n +1          
  

Page No 506:

Question 34:

How many terms of the AP 3, 7, 11, 15, ... will make the sum 406?
(a) 10
(b) 12
(c) 14
(d) 20

Answer:

(c) 14
Here, = 3 and d = (7-3) = 4
Let the sum of n terms be 406 .
Then,
we have:
 Sn = n22a + n-1d = 406    n22×3 + n-1×4 = 406     n3 + 2n -2  = 406      2n2 +n - 406 = 0     2n2 -28n +29n - 406 = 0     2n (n -14) + 29 (n -14) = 0      (2n +29) (n -14) =0    n = 14                       (n can't be a fraction)   
  
Hence, 14 terms will make the sum 406.

Page No 506:

Question 35:

The sam of the first 100 natural numbers is
(a) 4950
(b) 5050
(c) 5000
(d) 5150

Answer:

(b) 5050


The first 100 natural numbers are 1, 2, 3, 4, ...,100.
Here, = 1 and d = (2 - 1) = 1
 
The sum of n terms of an AP is given bySn=n22a+n-1d S100 = 1002×[2×1+100-1×1]        = 50×[2+99]=50×101=5050

Page No 506:

Question 36:

The sum of all odd numbers between 100 and 200 is
(a) 3750
(b) 2352
(c) 2450
(d) 2468

Answer:

(d) 7500

All odd numbers between 100 and 500 are 101, 103, 105, ..., 199.
This is an AP in which a = 101, d = (103 - 101) = 2 and l = 199.
Let the number of terms be n.
Then, Tn = 199
a + (n-1d = 199
⇒ 101 + (n - 1​) ⨯​ 2 = 199
⇒ 2n = 100
⇒ n = 50

∴ Required sum =  n2a+l
                     =502101 + 199 = 25 ×300 = 7500
  Hence, the required sum is 7500.

Page No 506:

Question 37:

The sum of all even natural numbers less than 100 is
(a) 2272
(b) 2352
(c) 2450
(d) 2468

Answer:

(c) 2450

All the even natural numbers less than 100 are 2, 4, 6, 8, ..., 98.
This is an AP in which a = 2, d = (4 - 2) = 2 and  l = 98
Let the number of terms be n.
Then Tn = 98
a + (n-1d = 98
⇒ 2 + ( n - 1​) ⨯​ 2 = 98
⇒ 2n = 98
⇒ n = 49

Required sum =   n2a+l
                     =4922 + 98 = 49 ×50 = 2450
  Hence, the required sum is 2450.

Page No 506:

Question 38:

The sum of the first fifteen multiples of 8 is
(a) 840
(b) 960
(c) 872
(d) 1081

Answer:

(b) 960

The first 15 multiples of 8 are 8,16,24, 32, ..., 120.
This is an AP in which a = 8, d = (16 - 8) = 8, l = 120 and n =  15
∴ Required sum =  n2a+l
                      =  152×8 + 120 = 15×64 = 960
                   
Hence, the required sum is 960.

Page No 506:

Question 39:

In an AP, the first term is 22, nth term is −11 and the sum of first n terms is 66. The value of n is
(a) 10
(b) 12
(c) 14
(d) 16

Answer:

(b) 12

Here, a = 22, Tn = -11 and Sn = 66
Let d be the common difference of the given AP. 
Then Tn = -11 
⇒ a + (n - 1)d = 22 + (n-1)d = -11
⇒ (n - 1)d = -33         ...(i)

The sum of n terms of an AP is given by
Sn =  n22a + n-1d = 66        n22×22 + (-33)  = n2 ×11= 66        n = 12        [Substituting the value of (n - 1)d from (i)]

Page No 506:

Question 40:

In an AP, the first term is 8, nth term is 33 and the sum of first n terms is 123. Thn, d = ?
(a) 5
(b) −5
(c) 7
(d) 3

Answer:

(a) 5

​​Here, a = 8, Tn = 33 and Sn = 123
Let d be the common difference of the given AP. 
Then, Tn = 33 
⇒ a + (n - 1)d = 8 + (n-1)d =  33
⇒ (n - 1)d = 25               ...(i)

The sum of terms of an AP is given by
Sn =  n22a + n-1d = 123        n22×8 + 25  = n2 ×41= 123        n = 6        [Substituting the value of (n - 1)d from (i)]
Putting the value of n in (i), we get:
5d = 25
d = 5

Page No 506:

Question 41:

The sum of n terms of an AP is given by Sn=(2n2+3n). What is the common difference of the AP?
(a) 3
(b) 4
(c) 5
(d) 9

Answer:

( b) 4
Given: Sn = (2n2  + 3n)              ...(i)
Replacing n by (n - 1) in (i), we get:
Sn-1 = 2(n - 1)2 + 3(n - 1)
        = 2(n2 - 2n + 1) + 3n - 3
        = 2n2 - n - 1
  ∴​ Tn = ( Sn - Sn-1)
​    = 
(2n2 + 3n) - (2n2 - n - 1) =  4n + 1
∴ nth term, Tn = (
4n + 1)       ...(ii)

Putting n = 1 in (ii), we get:
T1 = (4 ⨯ 1) + 1 = 5

 Putting n = 2 in (ii), we get:
T2 = (4 ⨯ 2) + 1 = 9
∴ Common difference, d = T2 - T1 = 9 - 5 = 4



Page No 507:

Question 42:

The sum of n terms of an AP is given by Sn=(3n2+5n). Which of its terms is 164?
(a) 25th
(b) 26th
(c) 27th
(d) 28th

Answer:

( c) 27th
Given: Sn = (3n2 + 5n)           ...(i)
Replacing n by (n - 1) in (i), we get:
Sn-1 = 3(n - 1)2 + 5(n - 1)
        = 3(n2 - 2n + 1) + 5n - 5
        = 3n2 - n - 2
  ∴​ Tn = (Sn - Sn-1)
​    = 
(3n2 + 5n) - (3n2 - n - 2 ) =  6n + 2
∴ nth term, Tn = (
6n + 2)                ...(ii)

Now,
Tn = 164
⇒ (
6n + 2) = 164
⇒ 6n = 162
n = 27
Hence, the 27th term of AP is 164.

Page No 507:

Question 43:

The sum of first p terms of an AP is (ap2+bp). What is the common difference of the AP?
(a) a
(b) 2a
(c) (a + b)
(d) 3a + b

Answer:

( b) 2a

Given: Sp = (ap2 + bp)          ...(i)
Replacing p by (p - 1) in (i), we get:
Sp-1 = a(p - 1)2 + b(p - 1)
        = a(p2 - 2p + 1) + bp - b
        = ap2 - 2ap + a + bp - b
  ∴​ Tp = ( Sp - Sp-1)
​             = 
(ap2 + bp) - (ap2 - 2ap + a + bp - b )
             =  2ap - a + b
∴ pth term, Tp = (
2ap - a + b)             ...(ii)

Putting p = 1 in (ii), we get:
T1= (2a ⨯ 1) - a + b = a + b

Putting p = 2 in (ii), we get:
T2 = (2a ⨯ 2) - a + b = 3a + b
∴ Common difference, d = T2 - T1 = 3a + b - a - b = 2a

Page No 507:

Question 44:

The famous mathematician associated with finding the sum of first 100 natural numbers, is
(a) Euclid
(b) Gauss
(c) Newton
(d) Pythagoras

Answer:

(b) Gauss is associated with finding the sum of the first 100 natural numbers.

Page No 507:

Question 45:

How many numbers lie between 10 and 300, which when divided by 4 leave a remainder 3?
(a) 71
(b) 72
(c) 73
(d) 74

Answer:

(c) 73

The required numbers are 11, 15, 19, ..., 299.
This is an AP in which a = 11, d = (15 - 11) = 4 and Tn = 299
Now, Tn = 299
a + (n - 1)d = 299
⇒ 11 + (n - 1) ⨯ ​4 = 299
⇒ 4n = 292
n = 73
Hence, 73 numbers, which when divided by 4 leave a remainder 3, lie in between 10 and 300.

Page No 507:

Question 46:

Two APs have the same common difference. Their first terms are −1 and −8, respectively. The difference between their 4th terms is
(a) −1
(b) −8
(c) 7
(d) −9

Answer:

(c) 7

Let the common difference of each AP be d. Then we have:
T4 = a + (n - 1)d = -1 + 3d
t4 = a + (n - 1)d = -8 + 3d
∴ (T4 - t4) = (-1 + 3d)(-8 + 3d) = 7
Hence, the difference between their 4th terms is 7.

Page No 507:

Question 47:

The sum of all multiples of 7 between 0 and 500 is
(a) 13916
(b) 17892
(c) 24353
(d) 16984

Answer:

(b) 17892

The multiples of 7 between 0 and 500 are 7, 14, 21, 28, ..., 497.
This is an AP in which a = 7, d = (14 - 7) = 7.
Let the number of terms be n.
Then Tn = 497 
a + (n - 1)d = 497
⇒ 7 + (n - 1) ⨯ ​7 = 497
⇒ 7n = 497
n = 71
 
∴ S17 =  n2a+l = 712×7 +497 = 71 ×252 = 17892 

Page No 507:

Question 48:

How many natural numbers are there between 23 and 100 which are exactly divisible by 6?
(a) 13
(b) 12
(c) 11
(d) 16

Answer:

(a) 13

The required numbers are 24, 30, 36, 42, ..., 96.
This is an AP in which a = 24, d = (30 - 24) = 6.
Let the number of terms be n.
Then Tn = 96 
a + (n - 1)d = 96
⇒ 24 + (n -1) ⨯ ​6 = 96
⇒ 6n = 78
n = 13

Hence, 13 numbers which are exactly divisible by 6 lie in between 23 and 100.

Page No 507:

Question 49:

The sum of all 2-digit numbers divisible by 5 is
(a) 1035
(b) 1245
(c) 1230
(d) 945

Answer:

(d) 945

The 2-digit numbers divisible by 5 are 10, 15, 20, 25, ..., 95.
This is an AP in which a = 10, d = (15 - 10) = 5.
Let the number of terms be n.
Then Tn = 95 
a + (n - 1)d = 95
⇒ 10 + (n - 1) ⨯ ​5 = 95
⇒ 5n = 90
n = 18

∴ S18n2a+l = 182×10+95 = 9×105 = 945

Page No 507:

Question 50:

Match the following columns

Column I Column II
(a) How many terms are there in
the AP 8, 13, 18, 23,..., 163?
(p) 65
(b) 12th term from the end of the
AP 9, 12, 15, 18, ..., 98 is
(q) 25
(c) The sum of the AP
5 + 12 + 19 + 26 + ... + 145 is
(r) 32
(d) The number of terms in the AP
(−5) + (−1) + 3 + 7 + ... + 91 is
(s) 1575

Answer:

(a) - (r), (b) -(p), (c) - (s), (d) - (q)

(a) a = 8, d = 5 and l = 163
Tn = a + (n − 1)d = 163
       ⇒ 8 + (n − 1) ⨯ 5 = 163

       ⇒ 5n = 160
      ⇒ n = 32

(b) l = 98 and d = 3
∴ 12th term from the end of the AP = [l − (n - 1)d]
                                                     = 98 - 11​ ⨯ 3
                                                     = 98 - 33 = 65

(c) Here, a = 5, d = 7 and l = 145
 Tn = 5 + (n - 1) ​⨯ 7 = 145
⇒ 7n = 147
 ⇒ n = 21
∴ S21n2a+l = 212×5 +145 = 21×75 = 1575

(d) Here, a = -5, d = 4 and l = 91
 ​ Tn = 91
⇒ -5 + (n - 1) ​⨯ 4 = 91
⇒ 4n = 100
n = 25



Page No 508:

Question 51:

Match the following columns:

Column I Column II
(a) In an AP, it is given that T5 = −9
and T9 = 7. Then, T14 = ?
(p) 8
(b) In an AP, Sn = (n2 + 3n). Then,
T16 = ?
(q) 34
(c) In an AP, T10 = 44 and T15 = 64.
Then, its first term is
(r) 15
(d) How many 2-digit numbers are
divisible by 6?
(s) 27

Answer:

(a) - (s), (b) - (q), (c) - (p ), (d) - ( r)


(a) T5 = a + 4d = −9      ...(i)
   T9 = a + 8d = 7           ...(ii)
Subtracting (i) from (ii), we get:
  4d = 16
 d = 4
On putting the value of d in (i), we get a = −25.
∴ T14 = a + 13d = −25 + 13 ⨯ 4 = −25 + 52 = 27


(b)  Sn = (n2 + 3n)     ...(i)
Replacing n by (n
1) in (i), we get;
Sn-1 = (n
1)2 + 3(n 1)
        = (n2 
2n + 1) + 3n  3
        = n2 + n 
2
  ∴​ Tn = (Sn 
 Sn-1)
​    = 
(n2 + 3n) (n2 + n  2) = 2n + 2
∴ nth term, Tn = (
2n + 2)                ...(ii)
  Putting n = 16 in (ii), we get:
  T16= (2 ⨯ 16) + 2 = 34
  Hence, the 16th term is 34.


(c) T10 = a + 9d = 44       ...(i)
   T15 = a + 14d = 64       ...(ii)
  Subtracting (i) from (ii), we get:
   5d = 20
 ∴ d = 4
 On putting the value of d in (i), we get a = 8.

(d) All the 2-digit numbers divisible by 6 are 12, 18, 24, 30, ..., 96.
    This is an AP whose first term is 12 and the common difference is 6.
    Let 96 be the nth term of the AP.   
Now, Tn = 96
a + (n − 1)d = 96
⇒ 12 + (n -1) 6  = 96  
⇒ (
n 1) = 14
n = 15

Page No 508:

Question 52:

Assertion (A)
8th term from the end of the AP 18, 14, 10, 6, ..., −30 is −2.

Reason (R)
The nth term from the end of an AP with last term l and common difference d is l −(n − 1)d.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R)is true.

Answer:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

Explanation:
Here, l =
30, n = 8 and d = 4
∴ 8th term from the end = l (n 1) ⨯ d = 30 7 ⨯ (4) = 30 + 28 = 2
Hence, Assertion (A) is true and clearly, Reason (R) gives Assertion (A).

Page No 508:

Question 53:

Assertion (A)
Sum of first n positive integers is given by Sn=12n(n+1).
Reason (R)
In an AP with first term a and common difference d, the sum of nth terms is given by Sn=n22a+(n-1)d.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R)is true.

Answer:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

Sn = 1+2+3+4+...+n      = n22×1+n-1×1=n2n+1 
This proves Assertion (A) and clearly, Reason (R) gives Assertion (A).



Page No 509:

Question 54:

Assertion (A)
If the first term of an AP is 6, last term is 62 and the sum of the given terms is 510. Then, there are 14 terms in the given AP.

Reason (R)
If a is the first term, l is the last term and n is the number of terms of an AP, then Sn=n2(a+1).

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R)is true.

Answer:

(d) Assertion (A) is false and Reason (R) is true.
 
The Reason (R) is clearly true.

Now, n2a+l=510 n26+62=510n=51034=15


Thus, Assertion (A) is false.



Page No 515:

Question 1:

Which term of the AP 2, −1, −4, −7, ... is −40?
(a) 8th
(b) 11th
(c) 14th
(d) 15th

Answer:

(d) 15th

 
Here, a = 2, d = (−1 − 2) = −3
Let the nth term be −40.
Then Tn = 40  
⇒ a + (n − 1)d = −40
⇒2 + (n − 1) ⨯ (−3) = −40
​​⇒ 3n = 45
​​⇒ n = 15

Page No 515:

Question 2:

Which term of the AP 20, 17, 14, ... is the first negative term?
(a) 6th
(b) 7th
(c) 8th
(d) 9th

Answer:

( c) 8th

Here, a  = 20 and d = (17 − 20) = −3
Let the nth term of the given AP be the first negative term.
Then
Tn < 0
 ⇒ 
a + (n −1)d < 0
 ⇒ 20 + (n1) ⨯ (−3) < 0
 ⇒ 23 − 3n < 0
 ⇒ 23 < 3n

 ⇒ n > 7.6
∴ n = 8
Hence, the 8th term of the given AP is the first negative term.

Page No 515:

Question 3:

What is the sum of all natural numbers from 1 to 100?
(a) 4550
(b) 4780
(c) 5050
(d) 5150

Answer:

(c) 5050
 

All the natural numbers from 1 to 100 are 1, 2, 3, 4, ..., 100.
Here, = 1, n = 100 and l = 100
∴ Sn =n2a+l=10021+100=50×101=5050

Page No 515:

Question 4:

If a, (a − 2), 3a are in AP, then a = ?
(a) −3
(b) −2
(c) 2
(d) 3

Answer:

(b) 2

If a, (a 2) and 3a are in AP, then we have:
(a 2) a = 3a (a 2)
⇒ 2a = 4
a = 2

Page No 515:

Question 5:

How many terms are there in the AP 7, 11, 15, ..., 139?

Answer:

Here, a = 7, d = (11 7) = 4 and l = 139
Let there are n terms in the given AP.
Then Tn = 139
a + (n 1)d = 139
​⇒ 7 + (n 1) ⨯ ​4 = 139
⇒ 4n = 136
n = 34
Hence, there are 34 terms in the AP.

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Question 6:

Is 51 a term of 5, 8, 11, 14, ...,?

Answer:

Here, a = 5, d = (8 5) = 3 
Let the nth term of the given AP be 51.
Then Tn = 51
a + (n 1)d = 51
​⇒ 5 + (n 1) ⨯ ​3 = 51
⇒ 3n = 49
⇒ n = 1613, but n cannot be a fraction.
∴ 51 is not a term of the given AP.

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Question 7:

Which term of the AP 18, 15, 12,... is 0?

Answer:

In the given AP, first term, a = 18 and common difference, d = (15 − 18) = −3.

Let its nth term be 0. Then we have:
Tn = 0  
⇒ 
a + (n − 1)d = 0
⇒ 18 + (n − 1) ⨯ (−3) = 0
⇒ 21 − 3n = 0
⇒ 3n = 21
n = 7
Hence, 0 is the  7th term of the given AP.

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Question 8:

If (5x + 2), (4x − 1) and (x + 2) are in AP, then x = ?

Answer:

Let (5x + 2), (4x 1) and (x + 2) are consecutive terms of an AP.
Then we have:
(4x 1)  ​(5x + 2) = ​(x + 2)  ​(4x 1)
⇒ (x 3) = (3x + 3)
⇒ 2x = 6
x = 3
∴ If (5x + 2), (4x − 1) and (x + 2) are in AP, then x = 3.

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Question 9:

If Sn=(3n2+2n), find the first term and the common difference of the AP.

Answer:

Given: 
Sn = (3n2 + 2n)            ...(i)
Replacing n by (n
1) in (i), we get:
Sn-1 = 3(n
1)2 + 2(n 1)
        = 3(n2 
2n + 1) + 2n  2
        = 3n2 
4n + 1
  ∴​ Tn = ( Sn 
 Sn -1)
​    = 
(3n2 + 2n) − (3n2 − 4n + 1) = 6n − 1
∴ nth term, Tn = (
6n − 1)       ...(ii)

Putting n = 1 in (ii), we get:
 T1 = (6 ⨯ 1) − 1 = 5

Putting n = 2 in (ii), we get:
T2 = (6 ⨯ 2) 1 = 11
∴ Common difference, d = T2  T1 = 11 5 = 6
Thus, a = 5 and d = 6

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Question 10:

Find the sum of 12 terms of the AP 5, 8, 11, 14,...

Answer:

Here, a = 5, = (8 5) = 3 and n = 12

Using the formula, Sn = n22a + n-1d, we get:
S12 = 1222×5 + 12-1 ×3                [ a = 5, d = 3 and n= 12]       = 6×10 + 33 = 6 ×43 = 258

Hence, the sum of the first 12 terms of the given AP is 258.

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Question 11:

In an AP, it is given that T8 = 31 and T15 = 45. Find the AP.

Answer:

Let the first term of the given AP be a and the common difference be d.
Then we have:
Tn = a + (n 1)d
T8 = a + 7d = 31     ...(i)
T15+ 14d = 45    ...(ii)
Subtracting (i) from (ii), we get:
7d = 14
d = 2
On putting the value of d in (i), we get:
a = 31 14 = 17
∴ The given AP is 17, 19, 21, 23, ...

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Question 12:

In an AP, it is given that Tn = (2n + 1). Find the sum of its n terms.

Answer:

We have:
Tn = (2n + 1)
So, T1= 2 ⨯ 1 + 1 = 3,
T2 = ​2 ⨯ 2 + 1 = 5
Now, d = T2 T1
           = 5 3 =  2
Thus, a = 3 and d =  2
The sum of n terms of an AP is given by

Sn = n22a + n-1d      = n22×3 + n-1×2  = n [ 3 + n -1] = nn+2

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Question 13:

Find the sum 25 + 28 + 31 + .+ 100.

Answer:

Here, a = 25, d = (28 25) = 3 and l = 100
Let there be n terms in the given AP.
Then Tn = 100
     ⇒ a + (n 1)d = 100
     ⇒ 25 + (n 1)⨯​3 = 100
     ⇒ 3n = 78
     ⇒ n = 26

Sum of n terms of an AP is given by
Sn = n2a +l  = 26225 + 100 =13×125 = 1625     

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Question 14:

Verify that −130 is a term of the AP 11, 8, 5, 2,.... . Which term is this?

Answer:

Here, a = 11, d = (8 − 11) = −3 
Let the nth term of the given AP be −130.
Then
Tn = −130
 ⇒ a + (n − 1)d = −130
 ​⇒ 11 + (n −1) ⨯ ​(−3) = −130
⇒ 3n = 144
⇒ n = 48
Thus, −130 is the 48th term of the given AP.

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Question 15:

Find the sum of all 2-digit numbers divisible by 6.

Answer:

The first 2-digit number divisible by 6 is 12 and the largest 2-digit number divisible by 6 is 96.
Thus, the AP is 12, 18, 24, 30, ..., 96, where 96 is the last term of the AP.
Let it be the nth term of the AP, where a = 12, d = (18 − 12) = 6 and l = 96
Then Tn = 96
    ⇒  a + (n -1) d = 96
    ⇒ 12 + (n -1​) ⨯​ 6 =96 
    ⇒  6n = 90
    ⇒n = 15
Hence, there are 15 terms in the AP.

Required sum = n2a+l
                      = 15212 + 96 = 15×54 = 810
Hence, the sum of all the 2-digit numbers divisible by 6 is 810.

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Question 16:

If a = 8, Tn = 63 and Sn = 210, find the value of n.

Answer:

Here, a =  8, Tn = 62 and Sn = 210
Now, a + (n − 1)d = 62

⇒ 8 + (n  1)d = 62
⇒ (
n 1)d = 54

Now, Sn = n22a + (n-1)d = 210               As (n -1) d = 55    n22×8 + 54 = 210    70 n = 210×2     n = 6           
 

 

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Question 17:

How many terms of the AP −15, −11, −7, ... are needed to make the sum 744?

Answer:

Here, a  = −15 and  d = [−11 − (−15)] =  4
Let the required number of terms be n. Then we have: 
 
   Sn =744  n22a + n-1d = 744n22×(-15) + n-1×4 = 744n2-30 +4n -4 = 744 n [ 2n -17] = 7442n2 - 17n - 744= 0 2n2 - 48n  +31 n - 744= 02n  (n -24) + 31( n - 24) = 0( 2n +31) ( n -24) = 0n = 24 or n = -31/2   ( n =-312 rejected, as n can't be a fraction)
Hence, 24 terms are needed to make the sum of 744.



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Question 18:

Find x, if −4 + (−1) + 2 + ... + x = 437.

Answer:

Here, a  = -4 and d = [−1 − (−4)] =  3
Let x be the nth term of the AP.
Sum of n terms of an AP is given by
 
   Sn=n22a + n-1d n22×(-4)+n-1×3=437n2-8+3n-3=437 n[3n-11]=437×23n2-11n-874=0 3n2-57n+46 n-874=03n(n-19)+46( n-19)=0(3n+46)( n-19)=0n=19 or n=-463   (n=-463 is rejected, as n can't be a fraction)
Thus, the sum of 19 terms is equal to 437.
x = T19​ = a + (n −1)d
a + 18d 
⇒ 
(−4) + 18 ⨯ 3 = 50

Hence, x = 50

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Question 19:

Find the sum of all multiples of 9 lying between 300 and 700.

Answer:

All multiples of 9 lying between 300 and 700 are 306, 315, 324, 333, ..., 693.

This is an AP in which a = 306, d = (315 − 306) = 9 and l = 693.
Let the number of terms be n. Then, we have:
Tn = 693 
a + (n 1) d = 693
⇒ 306 + ( n 1​) ⨯​ 9 = 693
⇒  9
n = 396
⇒ n = 44

∴ Required sum =  n2a+l
                     = 442×306 + 693 = 22×999 = 21978
                  
Hence, the required sum is 21978.

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Question 20:

The first and last terms of an AP are 4 and 81 respectively. If the common difference is 7, how many terms are there in the AP and what is thir sum ?

Answer:

Here, a  = 4, d = 7 and l = 81

Let its nth term be 81.
Then Tn = 81 
a + (n-1d = 81
⇒ 4 + ( n -1​) ⨯​ 7 = 81
⇒  7n = 84
⇒ n = 12
Hence, there are 12 terms in the AP.

∴ Snn2a+l
          = 122×4+81 = 6×85 = 510

Hence, the required sum is 510.



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