RS Aggarwal 2015 Solutions for Class 10 Math Chapter 11 - Arithmetic Progression

Answer:

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

Answer:

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

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

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

Answer:

The given AP is $6, 7\frac{3}{4}, 9\frac{1}{2}, 11\frac{1}{4},...$ 
First term, a = 6 and common difference, d $= 7\frac{3}{4} - 6 \Rightarrow \frac{31}{4} -6 \Rightarrow \frac{31 - 24}{4} = \frac{7}{4}$
$$\begin{gathered} \mathrm{Now}, T_{37} = a+(37-1) d = a+36d \\ =6+36\times \frac{7}{4}=6+63=69 \\ \therefore {37}^{th} \mathrm{term}=69 \end{gathered}$$

Answer:

   The given AP is $5, 4\frac{1}{2}, 4, 3\frac{1}{2}, 3,...$ 
    First term = 5
    Common difference$= 4\frac{1}{2} - 5 \Rightarrow \frac{9}{2} -5 \Rightarrow \frac{9 - 10}{2} = -\frac{1}{2}$
    $$\begin{gathered} \therefore a =5 \mathrm{and} d=-\frac{1}{2} \\ \mathrm{Now}, T_{25 }= a+(25-1)d=a+24d \\ =5+24\times \left(-\frac{1}{2}\right)=5-12=-7 \\ \therefore {25}^{th} \mathrm{term}=-7 \end{gathered}$$

Answer:

In the given AP, a  = 6 and d = (10 - 6) = 4
Suppose that there are n terms in the given AP.
Then T_n = 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.
             

Answer:

In the given AP, a  = 41 and d = (38 - 41) = -3
Suppose that there are n terms in the given AP.
Then T_n = 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.

Answer:

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

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

Answer:

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

Answer:

In the given AP, first term = $\frac{5}{6}$ and common difference, d $=\left(1- \frac{5}{6} = \frac{1}{6}\right)$.
Let its n^th term be 3.

$$\begin{gathered} \mathrm{Now}, T_n=3 \\ \Rightarrow a + (n-1)d = 3 \\ \Rightarrow \frac{5}{6} + (n-1)\times \frac{1}{6} = 3 \\ \Rightarrow \frac{2}{3} + \frac{n}{6} = 3 \\ \Rightarrow \frac{n}{6} = \frac{7}{3} \\ \Rightarrow n = 14 \end{gathered}$$

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

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.

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

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

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

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

Answer:

In the given AP, let the first term be a and the common difference be d.
Then, T_n = a + (n - 1)d ​
Now, we have:
T₁₀ = a + (10 - 1)d 
⇒ a + 9d  = 52       ...(1)
T₁₃ = a + (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

Answer:

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

Answer:

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

Answer:

Here, a = 5 and d = (15 - 5) = 10
The 31^st 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 n^th term.
Then T_n = 435
​⇒ 5 + (n - 1) ⨯ 10 = 435
⇒ 10n = 440
⇒ n = 44
Hence, the 44^th term will be 130 more than its 31^st term.  

Answer:

Let the n^th term of the given progressions be t_n and T_n, 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 n_^th term is given by
t_n = a + (n - 1)d 
⇒ 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 n_^th term is given by
T_n = A + (n - 1)D 
⇒ 3 + (n - 1) ⨯ 7
⇒ 7n - 4
Now, t_n = ​T_n
 ⇒ 61 + 2n​ = 7n - 4​
⇒ 65 = 5n
⇒ n = 13
Hence, the 13^th terms of the AP's are the same.

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:
T_n = 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.

Answer:

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

Answer:

Answer:

Answer:

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

Answer:

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

Answer:

Answer:

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

Answer:

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

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 = $\pm$3
Thus, a = 5 and d = $\pm$3
 Hence, the required numbers are (2, 5 and 8) or ( 8, 5 and 2).

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 = $\pm$6
Thus, a = 9 and d = $\pm$6
 Hence, the required numbers are (3, 9 and 15) or (15, 9 and 3).

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 = $\pm$6

Thus, a = 1 and d = $\pm$6
 Hence, the required numbers are ( -5, 1 and  7) or ( 7, 1 and -5).

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 
⇒ d = $\pm$3
 Thus, a = 8 and d = $\pm$3
 Hence, the required parts of 24 are (5, 8,11) or (11, 8, 5).

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 = $\pm$3
 Thus, a = 7 and d = $\pm$3
 Hence, the required terms are ( 4, 7,10) or ( 10, 7, 4).

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 = 360^o
⇒ 4a = 360 
⇒ a = 90

Hence, the required angles of a quadrilateral are $\left(90-15\right)°, \left(90-5\right)°, \left(90+5\right)° \mathrm{and} \left(90+15\right)°; \mathrm{or} 75°, 85°, 95° \mathrm{and} 105°.$

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

Answer:

Here, a = 2, d = (7 - 2) = 5 and n = 19
Using the formula, $S_n = \frac{n}{2}\left[2a + \left(n-1\right)d\right]$, we get:
$$\begin{gathered} S_{19}=\frac{19}{2}\left[2\times 2+\left(19-1\right)5\right] [\because a = 2, d = 5 \mathrm{and} n= 19] \\ = \frac{19}{2}\left[4+90\right]=19\times 47=893 \end{gathered}$$
Hence, the sum of the first 19 terms of the given AP is 893.

Answer:

Here, a = 1, d = (3-1) = 2 and n = 26
Using the formula, $S_n = \frac{n}{2}\left[2a + \left(n-1\right)d\right]$, we get:
$$\begin{gathered} S_{26}=\frac{26}{2}\left[1\times 2+\left(26-1\right)2\right] [\because a = 1, d = 2 \mathrm{and} n= 26] \\ =13\times \left[2+50\right]=13\times 52=676 \end{gathered}$$

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

Answer:

Here, a = 9, d = (7 - 9) = -2 and n = 18
Using the formula, $S_n = \frac{n}{2}\left[2a + \left(n-1\right)d\right]$, we get:
$$\begin{gathered} S_{18} = \frac{18}{2}\left[2\times 9+\left(18-1\right)\times (-2)\right] [\because a = 9, d = -2 \mathrm{and} n= 18] \\ = 9\times \left[18-34\right]=9\times (-16)=-144 \end{gathered}$$

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

Answer:

Here, a = 5, d = (13 - 5) = 8 and l = 181
Let the total number of terms be n.
Then T_n = 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 = $\frac{n}{2}\left(a+l\right)$

                   = $\frac{23}{2}\left(5 + 181\right) = 23\times 93 = 2139$
Hence, the required sum is 2139.

Answer:

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

∴ Required sum = $\frac{n}{2}\left(a+l\right)$
                          = $\frac{20}{2}\left(5 + 81\right) = 10\times 86 = 860$
Hence, the required sum is 860.

Answer:

In the given AP, first term, a  = 15 and common difference, d = (11 - 15) = -4.
Let its n^th term be -13.
Then T_n = -13 
⇒ a + (n - 1)d = -13
              ⇒ 15 + ( n - 1​) ⨯​ (-4) = -13
              ⇒ 4n = 32
              ⇒ n = 8
Hence, the 8^th term of the given AP is -13.
∴ Required sum = $\frac{n}{2}\left(a+l\right)$
                       = $\frac{8}{2}\left[15 + (-13)\right] = 8$
Hence, the required sum is 8.

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, T_n = 99 
⇒ a + (n - 1)d = 99
              ⇒ 12 + ( n - 1​) ⨯​ 3 = 99
              ⇒ 3n = 90
              ⇒ n = 30

∴ Required sum = $\frac{n}{2}\left(a+l\right)$
                         = $\frac{30}{2}\left[12 + 99\right] = 15\times 111 = 1665$
Hence, the required sum is 1665.

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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Answer:

Given: T_n = (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

Answer:

Given: S_n = (2n² + 5n)              ...(i)
Replacing n by (n - 1) in (i), we get:
S_n-1 = 2(n - 1)² + 5(n - 1)
        = 2(n² - 2n + 1) + 5n - 5
        = 2n² + n - 3
∴​ T_n = (S_n - S_n-1)
​        = (2n² + 5n) - (2n² + n - 3 ) =  4n +3
∴ n^th term, T_n = (4n +3)

Answer:

Given: S_n = (3n² - n)               ...(i)
Replacing n by (n - 1) in (i), we get:
S_n-1 = 3(n - 1)² - (n - 1)
        = 3(n² - 2n + 1) - n + 1
        = 3n² - 7n + 4
(i) Now,​ T_n = ( S_n - S_n-1)
​    = (3n² - n) - (3n² - 7n + 4 ) = 6n - 4
∴ n^th term, T_n = (6n - 4)                ...(ii)

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

Answer:

 $$\begin{gathered} S_n = \left(\frac{5n^2}{2}+\frac{3n}{2}\right) = \frac{1}{2}\left(5n^2 + 3n\right) ...\left(i\right) \\ \mathrm{Replacing} n \mathrm{by} (n-1) \mathrm{in} \left(\mathrm{i}\right), \mathrm{we} \mathrm{get}: \\ {S}_{n-1}=\frac{1}{2}\times \left[5{\left(n-1\right)}^2+3\left(n-1\right)\right] \\ =\frac{1}{2}\times \left[5n^2-10n+5+3n-3]\right]=\frac{1}{2}\times [5n^2 -7n+2] \\ \therefore T_n=S_n-S_{n-1} \\ =\frac{1}{2}\left(5n^2+3n\right)-\frac{1}{2}\times [5n^2 -7n+2] \\ =\frac{1}{2}\left(10n -2\right)=5n-1 ...\left(ii\right) \end{gathered}$$


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

Answer:

Let a be the first term and d be the common difference of the given AP.
Then we have:
$$\begin{gathered} S_n = \frac{n}{2}\left[2a + \left(n-1\right)d\right] \\ {S}_7 = \frac{7}{2}\left[2a + 6d\right] = 7[a +3d] \\ {S}_{17} =\frac{17}{2}\left[2a + 16d\right] = 17[a +8d] \end{gathered}$$

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:

Answer:

Let a be the first term and d be the common difference of the given AP.
Then we have:
$$\begin{gathered} S_n = \frac{n}{2}\left[2a + \left(n-1\right)d\right] \\ {S}_9 = \frac{9}{2}\left[2a + 8d\right] = 9[a +4d] \\ {S}_{20} =\frac{20}{2}\left[2a + 19d\right] = 10[2a +19d] \end{gathered}$$

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:

Answer:

Here, a = 4, d = 7 and l = 81
Let the n^th term be 81.
Then T_n = 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
$$\begin{gathered} S_n = \frac{n}{2}\left[a + l\right] \\ \therefore S_{12} = \frac{12}{2}\left[4 + 81\right] = 6\times 85=510 \end{gathered}$$

Answer:

Here, a = 22, T_n = -11 and S_n = 66
Let d be the common difference of the given AP. 
Then T_n = -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
$$\begin{gathered} S_n = \frac{n}{2}\left[2a + \left(n-1\right)d\right] = 66 \\ \Rightarrow \frac{n}{2}\left[2\times 22 + (-33)\right] = \left(\frac{n}{2}\right) \times 11= 66 \\ \Rightarrow n = 12 \end{gathered}$$        [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

Answer:

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

The sum of n terms of an AP is given by
$$\begin{gathered} S_n = \frac{n}{2}\left[a + l\right] = 155 \\ \Rightarrow \frac{n}{2}\left[2 + 29\right] = \left(\frac{n}{2}\right) \times 31= 155 \\ \Rightarrow n = 10 \end{gathered}$$        
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.

Answer:

On simplifying the given series, we get:

$$\begin{gathered} \left(1-\frac{1}{n}\right)+\left(1-\frac{2}{n}\right)+\left(1-\frac{3}{n}\right)+...n \mathrm{terms} \\ =(1+1+1+...n \mathrm{terms})-\left(\frac{1}{n}+\frac{2}{n}+\frac{3}{n}+...+\frac{n}{n}\right) \\ =n-\left(\frac{1}{n}+\frac{2}{n}+\frac{3}{n}+...+\frac{n}{n}\right) \\ \mathrm{Here}, \left(\frac{1}{n}+\frac{2}{n}+\frac{3}{n}+...+\frac{n}{n}\right) \mathrm{is} \mathrm{an} \mathrm{AP} \mathrm{whose} \mathrm{first} \mathrm{term} \mathrm{is} \frac{1}{\mathrm{n}} \mathrm{and} \mathrm{the} \mathrm{common} \mathrm{difference} \mathrm{is} \left(\frac{2}{n}-\frac{1}{n}\right)=\frac{1}{n}. \\ \mathrm{The} \mathrm{sum} \mathrm{of} n \mathrm{terms} \mathrm{of} \mathrm{an} \mathrm{AP} \mathrm{is} \mathrm{given} \mathrm{by} \\ Sn=\frac{n}{2}\left[2a+\left(n-1\right)d\right] \\ =n-\left[\frac{n}{2}\left\{2\times \left({\displaystyle \frac{1}{n}}\right)+(n-1)\times \left(\frac{1}{n}\right)\right\}\right] \\ =n-\left[\frac{n}{2}\left[\left(\frac{2}{n}\right)+\left(\frac{n-1}{n}\right)\right]\right]=n-\left\{\frac{n}{2}\left(\frac{n+1}{n}\right)\right\} \\ =n-\left(\frac{n+1}{2}\right)=\frac{n-1}{2} \end{gathered}$$

Answer:

(i) As the production of TV sets increases uniformly by a fixed number every year, the number of TV sets produced in the 1^st year, 2^nd year, 3^rd 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 n^th year be T_n.

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 1^st year is 2500.

(ii) Now, production of TV sets in the 8^th year is given by
    T₈ = a + 7d = 2500 + 7 ⨯ 1100 = 10200

(iii)  Sum of n terms of an AP is given by
       $$\begin{gathered} S_n=\frac{n}{2}\left[a+l\right] \\ \Rightarrow S_6=\frac{6}{2}\left[2500 + 8000\right]=3\times 10500=31500 \end{gathered}$$        
Thus, the total production of TV sets in the 6^th years is 31500.

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 +  a - 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.

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 S_n = 200.
The sum of n terms of an AP is given by
 $$\begin{gathered} S_n = \frac{n}{2}\left[2a + \left(n-1\right)d\right] \\ \mathrm{Here}, S_n = 200 \\ \therefore S_n =200 \\ \Rightarrow \left(\frac{n}{2}\right)\times \left[2\times 20 + \left(n-1\right)\times \left(-1\right)\right] = 200 \\ \Rightarrow \left(\frac{n}{2}\right)\times [ 40 - n +1] =200 \\ \Rightarrow n\times [41 - n] = 400 \\ \Rightarrow n^2 - 41n +400 = 0 \\ \Rightarrow n^2 - 25n - 16 n+400 =0 \\ \Rightarrow n(n - 25) - 16 (n - 25)=0 \\ \Rightarrow n = 25 or n = 16 \end{gathered}$$

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.

Answer:

If  $\frac{4}{5}$, a and 2 are three consecutive terms of an AP, then we have:
a - $\frac{4}{5}$ = 2 - a
⇒ 2a  = 2 + $\frac{4}{5}$
⇒ 2a = $\frac{14}{5}$
⇒ a = $\frac{7}{5}$​

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.

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.

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.

Answer:

We have:
T_n = (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.

Answer:

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

Answer:

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

Answer:

$$\begin{gathered} \mathrm{The} \mathrm{given} \mathrm{AP} \mathrm{is} \sqrt{8}, \sqrt{18}, \sqrt{32}, ... \\ \mathrm{On} \mathrm{simplifying} \mathrm{the} \mathrm{terms}, \mathrm{we} \mathrm{get}: \\ 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}, ... \\ \mathrm{Here}, a=2\sqrt{2} \mathrm{and} d=(3\sqrt{2}-2\sqrt{2})=\sqrt{2} \\ \therefore \mathrm{Next} \mathrm{term}, {\mathrm{T}}_4=a+3d=2\sqrt{2}+3\sqrt{2}=5\sqrt{2}=\sqrt{50} \end{gathered}$$

Answer:

:$$\begin{gathered} \mathrm{The} \mathrm{given} \mathrm{AP} \mathrm{is} \sqrt{2}, \sqrt{8}, \sqrt{18},... \\ \mathrm{On} \mathrm{simplifying} \mathrm{the} \mathrm{terms}, \mathrm{we} \mathrm{get}: \\ \sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, ... \\ \mathrm{Here}, a=\sqrt{2} \mathrm{and} d=(2\sqrt{2}-\sqrt{2})=\sqrt{2} \\ \therefore \mathrm{Next} \mathrm{term}, {\mathrm{T}}_4=a+3d=\sqrt{2}+3\sqrt{2}=4\sqrt{2}=\sqrt{32} \end{gathered}$$

Answer:

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

Answer:

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

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

Answer:

Answer:

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

Answer:

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

Answer:

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​

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
⇒ 12 - 2p = 5p - 16
⇒ 7​p = 28
⇒ p = 4

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

​

Answer:

(d) $\frac{7}{5}$

If $\frac{4}{5}$, a and 2 are three consecutive terms of an AP, then we have:
a - $\frac{4}{5}$ = 2 - a
⇒ 2a = 2 + $\frac{4}{5}$ 
⇒ 2a = $\frac{14}{5}$ 
⇒ a = $\frac{7}{5}$

Answer:

(d)  $\sqrt{32}$32−−√
$$\begin{gathered} \mathrm{The} \mathrm{given} \mathrm{AP} \mathrm{is} \sqrt{2}, \sqrt{8}, \sqrt{18}, ... \\ \mathrm{On} \mathrm{simplifying} \mathrm{the} \mathrm{terms}, \mathrm{we} \mathrm{get}: \\ \sqrt{2}, 2\sqrt{2}, 3\sqrt{2, }... \\ \mathrm{Here}, a=\sqrt{2} \mathrm{and} d=(2\sqrt{2}-\sqrt{2})=\sqrt{2} \\ \therefore \mathrm{Next} \mathrm{term}, {\mathrm{T}}_4=a+3d=\sqrt{2}+3\sqrt{2}=4\sqrt{2}=\sqrt{32} \end{gathered}$$