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Page No 61:

Question 1:

Which of the following sequences are A.P. ? If they are A.P. find the common difference .
(1) 2, 4, 6, 8,....
(2) 2, 52, 3, 73 ,....
(3) –10, –6, –2, 2...
(4) 0.3, 0.33, 0.333,...
(5) 0, –4, –8, –12,...
(6) -15,-15,-15,...
(7) 3, 3+2, 3+22, 3+32
(8) 127, 132, 137,...

Answer:

(1) 2, 4, 6, 8,....

The given sequence is an A.P.
Common difference = Second term − First term
                                 = 4 − 2
                                 = 2

(2) 2, 52, 3, 73 ,....

The given sequence is not an A.P.

(3) –10, –6, –2, 2...

The given sequence is an A.P.
Common difference = Second term − First term
                                 = (−6) − (−10)
                                 = 4

(4) 0.3, 0.33, 0.333,...

The given sequence is not an A.P.

(5) 0, –4, –8, –12,...

The given sequence is an A.P.
Common difference = Second term − First term
                                 = (−4) − (0)
                                 = −4

(6) -15,-15,-15,...

The given sequence is an A.P.
Common difference = Second term − First term
                                 = (−15) − (−15)
                                 = 0

(7) 3, 3+2, 3+22, 3+32

The given sequence is an A.P.
Common difference = Second term − First term
                                 = 3+2-3
                                 = 2

(8) 127, 132, 137,...

The given sequence is an A.P.
Common difference = Second term − First term
                                 = (132) − (127)
                                 = 5

Page No 61:

Question 2:

Write an A.P. whose first term is a and common difference is d in each of the following.
(1) a = 10, d = 5
(2) a = –3, d = 0
(3) a = –7, d = 12
(4) a = –1.25, d = 3
(5) a = 6, d = –3
(6) a = 19, d = –4

Answer:

(1) a = 10, d = 5

First term = a = 10
Second term = a + d = 10 + 5 = 15
Third term = a + 2d = 10 + 10 = 20
and so on...

Hence, the required A.P. is 10, 15, 20, 25, ...

(2) a = –3, d = 0

First term = a = –3
Second term = a + d = –3 + 0 = –3
Third term = a + 2d = –3 + 0 = –3
and so on...

Hence, the required A.P. is –3, –3, –3, ...

(3) a = –7, d = 12

First term = a = –7
Second term = a + d = -7+12=-132
Third term = a + 2d = –7 + 1 = –6
and so on...

Hence, the required A.P. is –7, –132, –6, ...

(4) a = –1.25, d = 3

First term = a = –1.25
Second term = a + d = –1.25 + 3 = 1.75
Third term = a + 2d = –1.25 + 6 = 4.75
and so on...

Hence, the required A.P. is –1.25, 1.75, 4.75, ...

(5) a = 6, d = –3

First term = a = 6
Second term = a + d = 6 – 3 = 3
Third term = a + 2d = 6 – 6 = 0
and so on...

Hence, the required A.P. is 6, 3, 0, ...

(6) a = 19, d = –4

First term = a = 19
Second term = a + d = 19 – 4 = 15
Third term = a + 2d = 19 – 8 = 11
and so on...

Hence, the required A.P. is 19, 15, 11, ...



Page No 62:

Question 3:

Find the first term and common difference for each of the A.P.
(1) 5, 1, –3, –7,...
(2) 0.6, 0.9, 1.2,1.5,...
(3) 127, 135, 143, 151,...
(4) 14,34,54,74,....

Answer:

(1) 5, 1, –3, –7,...

First term = 5
Common difference = Second term – First term
                                 = 1 – 5
                                 = –4

(2) 0.6, 0.9, 1.2,1.5,...

First term = 0.6
Common difference = Second term – First term
                                 = 0.9 – 0.6
                                 = 0.3

(3) 127, 135, 143, 151,...

First term = 127
Common difference = Second term – First term
                                 = 135 – 127
                                 = 8

(4) 14,34,54,74,....

First term = 14
Common difference = Second term – First term
                                 = 34-14
                                 = 12



Page No 66:

Question 1:

Write the correct number in the given boxes from the following A. P.
(i) 1, 8, 15, 22,...
Here a =    x   , t1 =    x   , t2 =    x   , t3 =    x   ,
t2 t1 =    x      x    =    x   
t3t2 =    x      x    =    x   d =    x   

(ii) 3, 6, 9, 12,...
Here  t1 =    x   , t2 =    x   , t3 =    x   , t4 =    x   ,
t2 t1 =    x   , t3t2 =    x      ∴ d =    x   

(iii) –3, –8, –13, –18,...
Here  t3 =    x   , t2 =    x   , t4 =    x   , t1 =    x   ,
t2 t1 =    x   , t3t2 =    x      ∴ a =    x   , d =    x   

(iv) 70, 60, 50, 40,...
Here  t1 =    x   , t2 =    x   , t3 =    x   ,...
a =    x   , d =    x   

Answer:

(i) 1, 8, 15, 22,...
Here a =    1   , t1 =    1   , t2 =    8   , t3 =    15   ,
t2 t1 =    8      1    =    7   
t3t2 =    15      8    =    7   d =    7   

(ii) 3, 6, 9, 12,...
Here  t1 =    3   , t2 =    6   , t3 =    9   , t4 =    12   ,
t2 t1 =    6-3=3x   , t3t2 =    9-6=3      ∴ d =    3   

(iii) –3, –8, –13, –18,...
Here  t3 =    -13    , t2 =    -8x   , t4 =    -18x   , t1 =    -3x   ,
t2 t1 =    -8--3=-5x   , t3t2 =    -13--8=-5x      ∴ a =    -3    , d =    -5x   

(iv) 70, 60, 50, 40,...
Here  t1 =    70   , t2 =    60   , t3 =    50   ,...
a =    70   , d =   -10x  

Page No 66:

Question 2:

Decide whether following sequence is an A.P., if so find the 20th term of the progression.
–12, –5, 2, 9, 16, 23, 30,...

Answer:

The given sequence is –12, –5, 2, 9, 16, 23, 30,...

Here,
First term (a) = a1 = –12
Second term = a2 = –5
Third term = a3 = 2

Common difference (d) = a2a1 = –5 – (–12) = 7
                                      = a3a2 = 2 – (–5) = 7

Since, a2a1 = a3a2

Hence, the given sequence is an A.P.

Now,
a20=a+n-1d      =-12+20-17      =-12+197      =121

Hence, the 20th term of the progression is 121.

Page No 66:

Question 3:

Given Arithmetic Progression 12, 16, 20, 24, . . . Find the 24th term of this progression.

Answer:

The given sequence is 12, 16, 20, 24, . . .

Here,
First term (a) = 12

Common difference (d) = a2a1 = 16 – (12) = 4

Now,
a24=a+n-1d      =12+24-14      =12+234      =104

Hence, the 24th term of the progression is 104.

Page No 66:

Question 4:

Find the 19th term of the following A.P.
7, 13, 19, 25,....

Answer:

The given sequence is 7, 13, 19, 25,....

Here,
First term (a) = 7

Common difference (d) = a2a1 = 13 – (7) = 6

Now,
a19=a+n-1d      =7+19-16      =7+186      =115

Hence, the 19th term of the progression is 115.

Page No 66:

Question 5:

Find the 27th term of the following A.P.
9, 4, –1, –6, –11,...

Answer:

The given sequence is 9, 4, –1, –6, –11,...

Here,
First term (a) = 9

Common difference (d) = a2a1 = 4 – (9) = –5

Now,
a27=a+n-1d      =9+27-1-5      =9+26-5      =-121

Hence, the 27th term of the progression is –121.

Page No 66:

Question 6:

Find how many three digit natural numbers are divisible by 5.
 

Answer:

The least positive three digit natural number divisible by 5 is 100.

The sequence divisible by 5 is 100, 105, 110, .... , 995.

Now,
an=a+n-1d995=100+n-15995=100+5n-5995=95+5n995-95=5n5n=900n=180

Hence, 180 three digit natural numbers are divisible by 5.

Page No 66:

Question 7:

The 11th term and the 21st term of an A.P. are 16 and 29 respectively, then find the 41th term of that A.P.

Answer:

It is given that,
a11 = 16 and a21 = 29

We know that,
an=a+n-1da11=a+11-1d16=a+10da=16-10d        ...1a21=a+21-1d29=a+20d29=16-10d+20d         from 129-16=10dd=1310               ...2Putting the value of d in 1, we geta=16-101310=3 a41=a+41-1d           =3+401310           =55

Hence, the 41th term of the A.P. is 55.

Page No 66:

Question 8:

11, 8, 5, 2,.... In this A.P. which term is number –151?

Answer:

The given sequence is 11, 8, 5, 2,.... –151.

Now,
an=a+n-1d-151=11+n-1-3-151=11-3n+3-151=14-3n3n=14+1513n=165n=55

Hence, 55th term is number –151.

Page No 66:

Question 9:

In the natural numbers from 10 to 250, how many are divisible by 4?

Answer:

The least positive natural number from 10 to 250 divisible by 4 is 12.

The sequence divisible by 4 is 12, 16, 20, .... , 248.

Now,
an=a+n-1d248=12+n-14248=12+4n-4248=8+4n248-8=4n4n=240n=60

Hence, 60 numbers are divisible by 4.

Page No 66:

Question 10:

In an A.P. 17th term is 7 more than its 10th term. Find the common difference.

Answer:

It is given that,
a17 = 7 + a10

We know that,
an=a+n-1da17=7+a10a+17-1d=7+a+10-1da+16d=7+a+9d16d=7+9d16d-9d=77d=7d=1

Hence, the common difference is 1.



Page No 72:

Question 1:

First term and common difference of an A.P. are 6 and 3 respectively ; find S27.

Answer:

It is given that,
First term (a) = 6
Common difference (d) = 3

We know that,
Sn=n22a+n-1d S27=27226+27-13            =27212+263            =27212+78            =272×90            =1215

Hence, S27 = 1215.

Page No 72:

Question 2:

Find the sum of first 123 even natural numbers.

Answer:

The given sequence is 2, 4, 6, 8 ...
First term (a) = 2
Common difference (d) = 2
Number of terms (n) = 123

We know that,
Sn=n22a+n-1d S123=12322(2)+123-12             =12324+1222             =12324+244             =1232×248             =15252

Hence, the sum of first 123 even natural numbers is 15252.

Page No 72:

Question 3:

Find the sum of all even numbers from 1 to 350.

Answer:

The given sequence is 2, 4, 6, 8 ... 350
First term (a) = 2
Common difference (d) = 2

Now,
an=a+n-1d350=2+n-12350=2+2n-22n=350n=175

Thus, the number of terms (n) = 175.

We know that,
Sn=n22a+n-1d S175=17522(2)+175-12             =17524+1742             =17524+348             =1752×352             =30800

Hence, the sum of all even numbers from 1 to 350 is 30800.

Page No 72:

Question 4:

In an A.P. 19th term is 52 and 38th term is 128, find sum of first 56 terms.

Answer:

It is given that,
a19 = 52
a38 = 128

Now,
an=a+n-1da19=a+19-1d52=a+18da=52-18d                 ...1a38=a+38-1d128=a+37d               ...2128=52-18d+37d          from 1128-52=19d19d=76d=4 a=52-184        =-20Thus, a=-20 and d=4.

We know that,
Sn=n22a+n-1d S56=5622(-20)+56-14            =28-40+220            =28180            =5040

Hence, the sum of first 56 terms is 5040.

Page No 72:

Question 5:

Complete the following activity to find the sum of natural numbers from 1 to 140 which are divisible by 4.

Sum of numbers from 1 to 140, which are divisible by 4 =         

Answer:

From 1 to 140, natural numbers divisible by 4 are 4, 8, ... 136.
a = 4
d = 4

Now,
tn=a+n-1d136=4+n-14136=4+4n-44n=136n=34

Thus, number of terms (n) = 34.

We know that,
Sn=n22a+n-1d S34=3422(4)+34-14            =178+132            =17140            =2380



Hence, the sum of numbers from 1 to 140, which are divisible by 4 =    2380    .

Page No 72:

Question 6:

Sum of first 55 terms in an A.P. is 3300, find its 28th term.

Answer:

It is given that,
S55 = 3300

We know that,
Sn=n22a+n-1dS55=5522a+55-1d3300=5522a+54d3300×255=2a+27d120=2a+28-1d60=a+28-1da28=60

Hence, its 28th term is 60.



Page No 73:

Question 7:

In an A.P. sum of three consecutive terms is 27 and their product is 504, find the terms.
( Assume that three consecutive terms in A.P. are ad , a , a + d .)

Answer:

Assume that three consecutive terms in A.P. are ad , a , a + d .

It is given that,
Sum of three consecutive terms = 27
Product of three consecutive terms = 504

a-d+a+a+d=273a=27a=9

a-d×a×a+d=5049-d×9×9+d=50481-d2=56d2=81-56d2=25d=±5

When d = 5,
The terms are 9, 14, 19.

When d = –5,
The terms are 9, 4, –1.

Page No 73:

Question 8:

Find four consecutive terms in an A.P. whose sum is 12 and sum of 3rd and 4th term is 14.

(Assume the four consecutive terms in A.P. are ad, a, a + d, a +2d)

Answer:

Assume that the four consecutive terms in A.P. are ad, a, a + d, a +2d .

It is given that,
Sum of four consecutive terms = 12
Sum of 3rd and 4th term = 14

a-d+a+a+d+a+2d=124a+2d=122a+d=62a=6-d           ...1

a+d+a+2d=142a+3d=146-d+3d=142d=14-62d=8d=4        from 12a=6-d2a=6-42a=2a=1

Hence, the terms are –3, 1, 5 and 9.

Page No 73:

Question 9:

If the 9th term of an A.P. is zero then show that the 29th term is twice the 19th term.

Answer:

It is given that,
a9 = 0

Now,
an=a+n-1da9=a+9-1d0=a+8da=-8d            ...1a29=a+29-1d      =-8d+28d      =20d               ...2a19=a+19-1d      =-8d+18d      =10d               ...3From 2 and 3, we geta29=2a19

Hence, the 29th term is twice the 19th term.



Page No 78:

Question 1:

On 1st Jan 2016, Sanika decides to save Rs 10, Rs 11 on second day, Rs 12 on third day. If she decides to save like this, then on 31st Dec 2016 what would be her total saving ?

Answer:

On 1st Jan 2016, Sanika saved Rs 10.
Next day Rs 11.
Third day Rs 12.

Her total savings = Rs (10 + 11 + 12 + ... + till 31st Dec)

Here,
a = 10
d = 1
n = 366 (since 2016 is a leap year)

Now,
Sn=n22a+n-1dS366=3662210+366-11             =18320+365             =70455

Hence, her total savings would be Rs 70455.

Page No 78:

Question 2:

A man borrows Rs 8000 and agrees to repay with a total interest of Rs 1360 in 12 monthly instalments. Each instalment being less than the preceding one by Rs 40. Find the amount of the first and last instalment.

Answer:

Money he borrows = Rs 8000
Total interest = Rs 1360

Total money he will pay after 12 months = Rs (8000 + 1360) = Rs 9360

It is given that,
Each instalment being less than the preceding one by Rs 40.

d = −40
n = 12
S12 = 9360

Now,
Sn=n22a+n-1dS12=1222a+12-1d9360=1222a+12-1-409360=62a-4402a-440=936062a-440=15602a=1560+4402a=2000a=1000

Also,
a12=a+12-1d      =1000+11-40      =1000-440      =560

Hence, the first and last instalment are Rs 1000 and Rs 560, respectively.

Page No 78:

Question 3:

Sachin invested in a national saving certificate scheme. In the first year he invested Rs 5000 , in the second year Rs 7000, in the third year Rs 9000 and so on. Find the total amount that he invested in 12 years.

Answer:

Sachin invested in first year = Rs 5000
Second year investment = Rs 7000
Third year investment = Rs 9000

Total investment = Rs (5000 + 7000 + 9000 + ..... + in 12 years)

Here,
a = 5000
d = 2000
n = 12

Now,
Sn=n22a+n-1dS12=1222a+12-1d       =12225000+12-12000       =610000+22000       =632000       =192000

Hence, the total amount that he invested in 12 years is Rs 192000.

Page No 78:

Question 4:

There is an auditorium with 27 rows of seats. There are 20 seats in the first row, 22 seats in the second row, 24 seats in the third row and so on. Find the number of seats in the 15th row and also find how many total seats are there in the auditorium ?

Answer:

Total number of rows = 27
Number of seats in first row = 20
Number of seats in second row = 22
Number of seats in third row = 24

Total number of seats = 20 + 22 + 24 + ..... + upto 27 rows

Here,
a = 20
d = 2
n = 27

Now,
Sn=n22a+n-1dS27=2722a+27-1d      =272220+262      =27240+52      =27292      =27×46      =1242

Hence, there are 1242 seats in the auditorium.

Also,
a15=a+15-1d      =20+142      =20+28      =48

Hence, the number of seats in the 15th row is 48.

Page No 78:

Question 5:

Kargil’s temperature was recorded in a week from Monday to Saturday. All readings were in A.P. The sum of temperatures of Monday and Saturday was 5°C more than sum of temperatures of Tuesday and Saturday. If temperature of Wednesday was –30° celsius then find the temperature on the other five days.

Answer:

The readings of the temperature were in A.P.
Let, first term = a
common difference = d

Temperature of Monday = a
Temperature of Tuesday = a + d
Temperature of Wednesday = a + 2d
Temperature of Thursday = a + 3d
Temperature of Friday = a + 4d
Temperature of Saturday = a + 5d

The sum of temperatures of Monday and Saturday was 5°C more than sum of temperatures of Tuesday and Saturday.
 a+a+5d=5+a+d+a+5d2a+5d=5+2a+6d6d-5d=-5d=-5

Also, the temperature of Wednesday was –30° celsius.
a+2d=-30a+2-5=-30a-10=-30a=-30+10a=-20

Hence, the temperature on the other five days were:
Monday = –20° celsius
Tuesday = –25° celsius
Thursday = –35° celsius
Friday = –40° celsius
Saturday = –45° celsius

Page No 78:

Question 6:

On the world environment day tree plantation programme was arranged on a land which is triangular in shape. Trees are planted such that in the first row there is one tree, in the second row there are two trees, in the third row three trees and so on. Find the total number of trees in the 25 rows.

Answer:

Total number of rows = 25
Number of trees in first row = 1
Number of trees in second row = 2
Number of trees in third row = 3

Total number of trees = 1 + 2 + 3 + ..... + upto 25 rows

Here,
a = 1
d = 1
n = 25

Now,
Sn=n22a+n-1dS25=2522a+25-1d      =25221+241      =2522+24      =25226      =25×13      =325

Hence, the total number of trees are 325.

Page No 78:

Question 1:

Choose the correct alternative answer for each of the following sub questions.
(1) The sequence –10, –6, –2, 2,...

(A) is an A.P., Reason d= –16 (B) is an A.P., Reason d= 4
(C) is an A.P., Reason d= –4 (D) is not an A.P.

(2) First four terms of an A.P. are ....., whose first term is –2 and common difference is –2.
(A)  –2, 0, 2, 4 (B) –2, 4, –8, 16
(C)  –2, –4, –6, –8 (D) –2, –4, –8, –16

(3) What is the sum of the first 30 natural numbers ?
(A) 464 (B) 465 (C) 462 (D) 461

(4) For an given A.P. t7 = 4, d = –4, n = 101, then a = ....
(A) 6 (B) 7 (C) 20 (D) 28

(5) For an given A.P. a = 3.5, d = 0, n = 101, then tn = ....
(A) 0 (B) 3.5 (C) 103.5 (D) 14.5

(6) In an A.P. first two terms are –3, 4 then 21st term is ...
(A) –143 (B) 143 (C) 137 (D) 17

(7) If for any A.P. d = 5 then t18 – t13 = ....
(A) 5 (B) 20 (C) 25 (D) 30

(8) Sum of first five multuiples of 3 is...
(A) 45 (B) 55 (C) 15 (D) 75

(9) 15, 10, 5,... In this A.P sum of first 10 terms is...
(A) –75 (B) –125 (C) 75 (D) 125

(10) In an A.P. 1st term is 1 and the last term is 20. The sum of all terms is = 399 then n = ....
(A) 42 (B) 38 (C) 21 (D) 19

Answer:

(1) The given sequence is  –10, –6, –2, 2,...

Here,
First term (a) = a1 = –10
Second term = a2 = –6
Third term = a3 = –2

Common difference (d) = a2a1 = –6 – (–10) = 4
                                      = a3a2 = –2 – (–6) = 4

Since, a2a1 = a3a2

Thus, the given sequence is an A.P.

Hence, the correct option is (B).


(2) It is given that,
First term (a) = –2
Common difference (d) = –2

Second term = a + d = –2 + (–2) = –4
Third term = a + 2d = –2 + 2(–2) = –6
Fourth term = a + 3d = –2 + 3(–2) = –8

Thus, first four terms of the A.P. are –2, –4, –6, –8

Hence, the correct option is (C).

(3) The given series is 1 + 2 + 3 + ... + 30

Here,
a = 1
d = 1
n = 30

Sn=n22a+n-1dS30=3022a+30-1d       =30221+291       =152+29       =15×31       =465

Hence, the correct option is (B).

(4) It is given that,
t7 = 4
d = –4
n = 101

Now,
tn=a+n-1dt7=a+7-1d4=a+6-44=a-24a=4+24a=28

Hence, the correct option is (D).

(5) It is given that,
a = 3.5
d = 0
n = 101

tn=a+n-1d   =3.5+n-10   =3.5

Hence, the correct option is (B).

(6) It is given that,
a = –3
a2 = 4

We know that,
a2=a+2-1d4=-3+dd=7

Now,
a21=a+21-1d      =-3+207      =-3+140      =137

Hence, the correct option is (C).

(7) It is given that,
d = 5

Now,
tn=a+n-1dt18-t13=a+18-1d-a+13-1d             =a+17d-a+12d             =5d             =55             =25

Hence, the correct option is (C).

(8) The given sequence is 3, 6, 9,...
Here,
a = 3
d = 3
n = 5

We know that,
Sn=n22a+n-1dS5=522a+5-1d     =5223+43     =526+12     =5218     =59     =45

Hence, the correct option is (A).

(9) The given sequence is 15, 10, 5,...
Here,
a = 15
d = –5

We know that,
Sn=n22a+n-1dS10=1022a+10-1d       =5215+9-5       =530-45       =5-15       =-75

Hence, the correct option is (A).

(10) It is given that,
First term (a) = 1
Last term (tn) = 20
Sum of terms (Sn) = 399

We know that,
tn=a+n-1dSn=n22a+n-1dSn=n2a+a+n-1dSn=n2a+tn399=n21+20399=21n221n=399×2n=79821n=38

Hence, the correct  option is (B).



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Question 2:

Find the fourth term from the end in an A.P. –11, –8, –5,...., 49.

Answer:

The given sequence is –11, –8, –5,...., 49.

Here,
First term (a) = –11
Common difference (d) = 3
n = 4
Last term (l) = 49

We know that,
nth term from the end=l-n-1d4th term from the end=l-4-1d                                     =49-33                                     =49-9                                     =40

Hence, the fourth term from the end is 40.

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Question 3:

In an A.P. the 10th term is 46 sum of the 5th and 7th term is 52. Find the A.P.

Answer:

It is given that,
t10 = 46
t5 + t7 = 52

Now,
tn=a+n-1dt10=a+10-1d46=a+9da=46-9d       ...1t5+t7=52a+5-1d+a+7-1d=52a+4d+a+6d=522a+10d=52246-9d+10d=52       from192-18d+10d=5292-8d=528d=92-528d=40d=5a=46-95       from1a=46-45a=1

Hence, the given A.P. is 1, 6, 11, 16, ....

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Question 4:

The A.P. in which 4th term is –15 and 9th term is –30. Find the sum of the first 10 numbers.

Answer:

It is given that,
t4 = –15
t9 = –30

Now,
tn=a+n-1dt4=a+4-1d-15=a+3da=-15-3d       ...1t9=a+9-1d-30=a+8da+8d=-30-15-3d+8d=-30       from1-15+5d=-305d=-30+155d=-15d=-3a=-15-3-3       from1a=-15+9a=-6

Hence, the given A.P. is –6, –9, –12, ....

Now,
Sn=n22a+n-1dS10=1022a+10-1d      =52-6+9-3      =5-12-27      =5-39      =-195

Hence, the sum of the first 10 numbers is –195.

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Question 5:

Two A.P.’ s are given 9, 7, 5, . . . and 24, 21, 18, . . . . If nth term of both the progressions are equal then find the value of n and n th term.

Answer:

The given sequence is
9, 7, 5, . . .

Here,
a = 9
d = –2

We know that,
an=a+n-1d    =9+n-1-2    =9-2n+2    =11-2n        ...1

Another given sequence is
24, 21, 18, . . . .

Here,
a = 24
d = –3

We know that,
a'n=a+n-1d    =24+n-1-3    =24-3n+3    =27-3n        ...2

It is given that,
nth term of both the progressions are equal.

From (1) and (2), we get
11-2n=27-3n3n-2n=27-11n=16

Hence, n = 16.

Now,
a16=11-216           from 1      =11-32      =-21

Hence, the value of n and nth term is 16 and –21, respectively.

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Question 6:

If sum of 3rd and 8th terms of an A.P. is 7 and sum of 7th and 14th terms is –3 then find the 10 th term .

Answer:

It is given that,
a3 + a8 = 7
a7 + a14 = –3

Now,
an=a+n-1da3+a8=7a+3-1d+a+8-1d=7a+2d+a+7d=72a+9d=72a=7-9d       ...1a7+a14=-3a+7-1d+a+14-1d=-3a+6d+a+13d=-32a+19d=-37-9d+19d=-3           from 17+10d=-310d=-3-710d=-10d=-12a=7-9-1            from 12a=7+92a=16a=8

Now,
a10=a+10-1d      =8+9-1      =8-9      =-1

Hence, the 10th term is –1.

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Question 7:

In an A.P. the first term is –5 and last term is 45. If sum of all numbers in the A.P. is 120, then how many terms are there ? What is the common difference ?

Answer:

It is given that,
a = –2
l = 45
Sn = 120

Now,
Sn=n2a+l120=n2-5+45120=n240120×2=n40240=n40n=24040n=6

Hence, there are 6 terms.

Also,
l=a+6-1d45=-5+5d45+5=5d5d=50d=10

Hence, the common difference is 10.

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Question 8:

Sum of 1 to n natural numbers is 36, then find the value of n .

Answer:

It is given that,
a = 1
d = 1
Sn = 36

Now,
Sn=n22a+n-1d36=n221+n-1136×2=n2+n-172=nn+1n2+n-72=0n2+9n-8n-72=0nn+9-8n+9=0n+9n-8=0n=-9 or n=8n=8                  n-9

Hence, the value of n is 8.



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Question 9:

Divide 207 in three parts, such that all parts are in A.P. and product of two smaller parts will be 4623.

Answer:

Let the three numbers be ad, a and a + d.

According to the question,
a-d+a+a+d=2073a=207a=69

Also,
a-da=462369-d69=462369-d=46236969-d=6769-67=dd=2

Hence, the three numbers are 67, 69 and 71.

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Question 10:

There are 37 terms in an A.P., the sum of three terms placed exactly at the middle is 225 and the sum of last three terms is 429. Write the A.P.

Answer:

Let the first term be a and the common difference be d.

Since, the A.P. contains 37 terms, the middle term is 37+12th=19th term.

According to the question,
a18+a19+a20=225a+18-1d+a+19-1d+a+20-1d=2253a+17d+18d+19d=2253a+54d=2253a+18d=225a+18d=2253a+18d=75a=75-18d             ...1

Also,
a35+a36+a37=429a+35-1d+a+36-1d+a+37-1d=4293a+34d+35d+36d=4293a+105d=4293a+35d=429a+35d=429375-18d+35d=143             from 117d=143-7517d=68d=4a=75-184             from 1a=75-72a=3

Hence, the resulting A.P is 3, 7, 11, ....

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Question 11:

If first term of an A.P. is a, second term is b and last term is c, then show that sum of all terms is a+c b+c-2a2b-a .

Answer:

It is given that,
First term = a
a2 = b
l = c

Now,
an=a+n-1da2=a+2-1db=a+dd=b-a        ...1l=a+n-1dc=a+n-1b-a         from 1c-a=n-1b-an-1=c-ab-an=c-ab-a+1n=c-a+b-ab-an=c+b-2ab-a        ...2

Also,
Sn=n2a+l     =c+b-2ab-a2a+c           from 2     =c+b-2aa+c2b-a

Hence, the sum of all terms is a+c b+c-2a2b-a.

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Question 12:

If the sum of first p terms of an A.P. is equal to the sum of first q terms then show that the sum of its first (p + q) terms is zero. (p ≠ q)

Answer:

We know,
Sn=n22a+n-1d

According to the question,
Sp=Sqp22a+p-1d=q22a+q-1dp2a+p-1d=q2a+q-1d2ap+p-1pd=2aq+q-1qd2ap+p2d-pd=2aq+q2d-qd2ap-2aq=q2d-qd-p2d+pd2ap-q=dq2-p2+dp-q2ap-q=dq+pq-p+dp-q2ap-q=dq+pq-p+p-q2ap-q=d-q+pp-q+p-q2ap-q=dp-q1-q-p2a=d1-q-p                  pq2a=d1-q-p           ...1

Now,
Sp+q=p+q22a+p+q-1d          =p+q21-q-pd+p+q-1d                from 1          =p+q2d1-q-p+p+q-1          =0

Hence, the sum of its first (p + q) terms is zero.

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Question 13:

If m times the mth term of an A.P. is eqaul to n times nth term then show that the (m + n)th term of the A.P. is zero.

Answer:

We know,
an=a+n-1d

According to the question,

mam=nanma+m-1d=na+n-1dam+m-1md=an+n-1ndam+m2d-md=an+n2d-ndam-an=n2d-nd-m2d+mdam-n=dn2-m2+dm-nam-n=dm+nn-m+dm-nam-n=dm+nn-m+m-nam-n=d-m+nm-n+m-nam-n=dm-n1-m-na=d1-m-n                  mna=d1-m-n           ...1

Now,
am+n=a+m+n-1d          =1-m-nd+m+n-1d                from 1          =d1-m-n+m+n-1          =0

Hence, the (m + n)th term of the A.P. is zero.

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Question 14:

Rs 1000 is invested at 10 percent simple interest. Check at the end of every year if the total interest amount is in A.P. If this is an A.P. then find interest amount after 20 years. For this complete the following activity.

Answer:

It is given that,
Principal (P) = Rs 1000
Rate (R) = 10%
Simple interest (S.I.) = P×R×T100

Simple interest after an year = 1000×10×1100=Rs 100

Simple interest after 2 years = 1000×10×2100=Rs 200

Simple interest after 3 years = 1000×10×3100=Rs 300

Hence, the total interest amount is in A.P. i.e. 100, 200, 300,....

Here,
a = 100
d = 100

Now,
an=a+n-1da20=a+20-1d      =100+19100      =100+1900      =2000

Hence, the interest amount after 20 years is Rs 2000.



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