# Arithmetic Progressions : Exercise 5.2 (Mathematics NCERT Class 10th)

Q.1     Fill in the blanks in the following table, given that a is the first term, d the common difference and ${a_n}$ the nth term of the AP :

Sol.      Blanks may be filled as under :
(i) ${a_n} = a + \left( {n - 1} \right)d = 7 + \left( {8 - 1} \right)3$
$= 7 + 7 \times 3 = 7 + 21 = 28$

(ii) ${a_n} = a + \left( {n - 1} \right)d$
$\Rightarrow$ $0 = - 18 + \left( {10 - 1} \right)d$

$\Rightarrow$ $18 = 9d$
$\Rightarrow$ $d = {{18} \over 9} = 2$

(iii) ${a_n} = a + \left( {n - 1} \right)d$
$\Rightarrow$ $- 5 = a + \left( {18 - 1} \right)\left( { - 3} \right)$
$\Rightarrow$ $- 5 = a + 17\left( { - 3} \right)$
$\Rightarrow$ $- 5 = a - 51$
$\Rightarrow$ $a = - 5 + 51 = 46$

(iv) ${a_n} = a + \left( {n - 1} \right)d$
$\Rightarrow$ $3.6 = - 18.9 + \left( {n - 1} \right)2.5$
$\Rightarrow$ $3.6 + 18.9 = \left( {n - 1} \right)2.5$
$\Rightarrow$ $22.5 = \left( {n - 1} \right)2.5$
$\Rightarrow$ $n - 1 = {{22.5} \over {2.5}}$
$\Rightarrow$ $n - 1 = 9$
$\Rightarrow$ $n = 9 + 1 = 10$

(v) ${a_n} = a + \left( {n - 1} \right)d = 3.5 + \left( {105 - 1} \right) \times 0$
$= 3.5 + 0 = 3.5$

Q.2     Choose the correct choice in the following and justify :
(i) 30th term of the 10, 7, 4, ..... is
(a) 97              (b) 77               (c) – 77               (d) – 87

(ii) 11th term of the $- 3, - {1 \over 2},2\,...,\,\,is$
(a) 28               (b) 22              (c) – 38                 (d) $- 48{1 \over 2}$
Sol.       (i) Here, a = 10, d = 7 – 10 = – 3, n = 30
We know that   ${a_n} = a + \left( {n - 1} \right)d$
Therefore, ${a_{30}}$ = 10 + (30 – 1) (–3) = 10 + 29 (– 3)
= 10 – 87 = – 77
So, (C) is the correct choice.

(ii) Here, a = – 3, $d = - {1 \over 2} - \left( { - 3} \right)$
$= {1 \over 2} + 3 = {{ - 1 + 6} \over 2}$
$= {5 \over 2},\,n = 11$
We know that ${a_n} = a + \left( {n - 1} \right)d$
Therefore, ${a_{11}} = - 3 + \left( {11 - 1} \right){5 \over 2}$
$= - 3 + 10 \times {5 \over 2} = - 3 + 25 = 22$
So, (B) is the correct choice.

Q.3      In the following APs, find the missing terms in the boxes :
(i) 2,  , 26                                                      (ii) , 13, , 3

(iii) 5, $9{1 \over 2}$                        (iv) –4,  , 6

(v)  , 38, , –22.

Sol.        (i) Let 2, , 26, be a + (a + d) and (a + 2d)
Therefore, a = 2 and a + 2d = 26
$\Rightarrow$ 2 + 2d = 26
$\Rightarrow$ 2d = 26 – 2 = 24

$\Rightarrow$ $d = {{24} \over 2} = 12$
Thus the missing term =  a + d = 2 + 12 = 14

(ii) Let , 13, , 3 be a, a + d, a + 2d and  a + 3d
Therefore, a + d = 13                    ... (1)
and, a + 3d = 3                                 ... (2)
(2) – (1) gives, 2d = – 10
$\Rightarrow$ d = – 5

Putting d = – 5 in (1), we get
a – 5 = 13
$\Rightarrow$ a = 13 + 5 = 18

Therefore, The missing terms are a, i.e., 18
and , a + 2d, i.e., 18 + 2 (– 5) = 18 – 10 = 8

(iii)  Let 5, $9{1 \over 2}$, be a, a +d, a +2d and a  + 3d .
Therefore, a = 5 ... (1)
and , $a + 3d = 9{1 \over 2} = {{19} \over 2}$                    ... (2)
(2) – (1) gives, $3d = {{19} \over 2} - 5 = {{19 - 10} \over 2} = {9 \over 2}$
$\Rightarrow$ $d = {1 \over 3} \times {9 \over 2} = {3 \over 2}$
Therefore, The missing terms are a + d, i.e.,
$5 + {3 \over 2} = 5 + 1{1 \over 2} = 6{1 \over 2}$
and a + 2d, 5 + 2 × ${3 \over 2}$ = 5 + 3 = 8

(iv) Let – 4, , 6 be a, a + d, a + 2 d, a + 3 d, a + 4 d and a + 5 d.
Therefore, a = – 4                                                              ... (1)
and, a + 5d = 6                                                                  ... (2)
(2) – (1) gives , 5d = 10 d = 2
Therefore, The missing terms are
a + d = – 4 + 2 = – 2
a + 2d = – 4 + 2 × 2 = – 4 + 4 = 0,
a + 3d = – 4 + 3 × 2 = – 4 + 6 = 2,
and, a + 4d = – 4+ 4 × 2 = – 4 + 8 = 4

(v) Let , 38, , – 22 be a, a + d, a + 2 d, a + 3 d, a + 4 d and a  + 5 d.
Therefore, a + d = 38 ... (1)
and a + 5d = – 22 ... (2)
(2) – (1) gives,  4d = – 60
$\Rightarrow$ d = – 15
Putting d = – 15 in (1), we get
a –15 = 38
$\Rightarrow$ 38 + 15 = 53
Therefore, The missing terms are a, i.e., 53.
a + 2d, i.e., 53 + 2 × (– 15 )= 53 – 30 = 23,
a + 3d, i.e., 53 + 3 × (– 15) = 53 – 45 = 8,
and a + 4 d, i.e., 53 + 4 × (– 15) = 53 – 60 = – 7

Q.4      Which terms of the AP : 3, 8, 13, 18, .... is 78 ?
Sol.         Clearly the given list is an AP.
We have, a = 3, d = 8 – 3 = 5
Let 78 be the nth term of the given AP then,
${a_n} = 78$
$\Rightarrow$ $a + \left( {n - 1} \right)d = 78$
Therefore, $3 + \left( {n - 1} \right)5 = 78$
$\Rightarrow$ $\left( {n - 1} \right)5 = 78 - 3$
$\Rightarrow$ $5\left( {n - 1} \right) = 75$
$\Rightarrow$ $n - 1 = 15$
$\Rightarrow$ $n = 15 + 1$
$\Rightarrow$ $n = 16$
Thus, 78 is the 16th term of the given list.

Q.5      Find the number of terms in each of the following APs :
(i) 7, 13, 19 ...., 205
(ii) 18, $15{1 \over 2}$, 13, ..., – 47
Sol.        (i) Clearly it forms an AP with first term a = 3 and common difference d = 13 – 7 = 6
(i) Let there be n terms in the given list. Then , nth term = 205.
$\Rightarrow$ a + (n –1) d = 205
$\Rightarrow$ 7 + (n –1) 6 = 205
$\Rightarrow$ 6 (n – 1) = 205 – 7
$\Rightarrow$ 6 (n – 1) = 198
$\Rightarrow$ $n - 1 = {{198} \over 6} = 33$
$\Rightarrow$ n = 33 + 1 = 34
Thus the given list contains 34 terms.

(ii) Let there be n terms in the given list.
18, $15{1 \over 2}$, 13, .... , – 47. Clearly, it forms an AP with first term a = 18 and common difference
$d = 15{1 \over 2} - 18$
$= {{31} \over 2} - 18 = {{31 - 36} \over 2} = {{ - 5} \over 2}$
Then, nth term = – 47.
$\Rightarrow$ $a + \left( {n - 1} \right)d = - 47$
$\Rightarrow$ $18 + \left( {n - 1} \right)\left( {{{ - 5} \over 2}} \right) = - 47$
$\Rightarrow$ $\left( {{{ - 5} \over 2}} \right)\left( {n - 1} \right)$ $= - 47 - 18 = - 65$
$\Rightarrow$ $n - 1 = - 65 \times {{ - 2} \over 5}$
$= - 13 \times - 2 = 26$
$\Rightarrow$ $n = 26 + 1 = 27$
There are 27 terms in the given AP.

Q.6      Check whether – 150 is a term of the AP : 11 , 8, 5, 2, ....
Sol.         Here, ${a_2} - {a_1} = 8 - 11 = - 3$
${a_3} - {a_2} = 5 - 8 = - 3$
${a_4} - {a_3} = 2 - 5 = - 3$
As ${a_{n + 1}} - {a_n}$ is same every time, so the given list of numbers is an AP.
Now , a = 11 , d = – 3
Let – 150 be the nth term of the given AP.
We know that ${a_n} = a + \left( {n - 1} \right)d$
$\Rightarrow$ $- 150 = 11 + \left( {n - 1} \right)\left( { - 3} \right)$
$\Rightarrow$ $- 3\left( {n - 1} \right) = - 150 - 11 = - 161$
$\Rightarrow$ $n - 1 = {{161} \over 3}$
$\Rightarrow$ $n = {{161} \over 3} + 1 = {{164} \over 3}$
But n should be a positive integer. So, our assumption was wrong and so - 150 is not a term of the given AP.

Q.7       Find the 31st term of an AP whose 11 th term is 38 and the 16th term is 73.
Sol.          Let a be the first term and d be the common difference.
Now, ${a_n} = a + \left( {n - 1} \right)d$
Therefore, ${a_{11}} = a + 10d = 38$                 ... (1)
and ${a_{16}} = a + 15d = 73$                              ... (2)
Subtracting (1) from (2) we get
$5d = 35\,$
$\Rightarrow$ $d = {{35} \over 5} = 7$
and then from (1),
a + 10 × 7 = 38
$\Rightarrow$ a = 38 – 70 = – 32
Therefore, ${a_{31}} = a + 30d = - 32 + 30 \times 7$
$= - 32 + 210 = 178$

Q.8       An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Sol.          Let a be the first term and d the common difference.
Now ${a_n} = a + \left( {n - 1} \right)d$
${a_3} = a + 2d = 12$ ... (1)
${a_{50}} = a + 49\,d = 106$ ... (2)
Subtracting (1) from (2) , we get -
$47d = 94$
$\Rightarrow$ $d = {{94} \over {47}} = 2$

and then from (1)
a + 2 × 2 = 12
$\Rightarrow$ a = 12 – 4 = 8

Therefore, ${a_{29}}$ = a + 28 d = 8 + 28 × 2
= 8 + 56 = 64

Q.9      If the 3rd and 9th terms of an AP are 4 and 8 respectively, which term of this AP is zero?
Sol.        Let a be the first term and d the common difference.
${a_3}$ = a + 2d = 4 ... (1)
${a_9}$ = a + 8d = – 8 ... (2)
Subtracting (1) from (2) , we get
6d = – 12 $\Rightarrow$ $d = {{ - 12} \over 6} = - 2$
and then from (1)
a + 2 × (– 2) = 4
$\Rightarrow$ a = 4 + 4 = 8
Let ${a_n} = 0$
$\Rightarrow$ a + (n – 1) d = 0
$\Rightarrow$ 8 + (n – 1) (– 2) = 0
$\Rightarrow$ (n – 1) (– 2) = – 8
$\Rightarrow$ $n - 1 = {{ - 8} \over { - 2}} = 4$
$\Rightarrow$ n = 4 + 1 = 5
Thus , 5th term is zero.

Q.10     The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Sol.         Let a be the first term and d the common difference.
It is given ${a_{17}} - {a_{10}} = 7$
$\Rightarrow$ (a + 16d) – (a + 9d) = 7 $\left[ {Since,{a_n} = a + \left( {n - 1} \right)d} \right]$
$\Rightarrow$ 7d = 7
$\Rightarrow$ d = 1
Thus, the common difference is 1

Q.11     Which term of the AP : 3, 15, 27, 39, ..... will be 132 more than its 54th term?
Sol.          Here, a = 3, d = 15 – 3 = 12. Then.
${a_{54}}$ = a + 53d = 3 + 53 × 12  = 3 + 636 = 639
Let ${a_n}$ be 132 more than its 54th term
i.e., ${a_n} = {a_{54}}$ + 132 = 639 + 132 = 771$\left[ {Since,{a_n} = a + \left( {n - 1} \right)d} \right]$
$\Rightarrow$ a + (n – 1)d = 771
$\Rightarrow$ 3 + (n – 1)12 = 771
$\Rightarrow$ 12(n – 1) = 771 – 3
$\Rightarrow$ 12(n– 1) = 768
$\Rightarrow$ $n - 1 = {{768} \over {12}} = 64$
$\Rightarrow$ n = 64 + 1 = 65
Thus, 65th term is 132 more than its 54th term.

Q.12      Two APs have the same common difference. The difference between their 100th term is 100, what is the difference between their 1000th terms?
Sol.            Let the two APs be ${a_1},{a_2},{a_3},....\,{a_n},...\,and\,{b_1},\,{b_2},\,{b_3}...,\,{b_n},...$
Also, let d be the common difference of two APs. Then,
${a_n} = {a_1} + \left( {n - 1} \right)d$
and ${b_n} = {b_1} + \left( {n - 1} \right)d$
$\Rightarrow$ ${a_n} - {b_n} = \left[ {{a_1} + \left( {n - 1} \right)d} \right]$
$- \left[ {{b_1} + \left( {n - 1} \right)d} \right]$
$\Rightarrow$ ${a_n} - {b_n} = {a_1} - {b_1}\,for\,all\,n\, \notin N$
$\Rightarrow$ ${a_{100}} - {b_{100}} = {a_1} - {b_1} = 100$ [Given]
Now, ${a_{1000}} - {b_{1000}} = {a_1} - {b_1}$
$\Rightarrow$ ${a_{1000}} - {b_{1000}} = 100$ [Since ${a_1} - {b_1} = 100$]
Hence, the difference between 1000th terms is the same as the difference between 100th terms, i.e., 100.

Q.13      How many three-digit numbers are divisible by 7 ?
Sol.          994 is the last 3-digit number divisible by 7. Thus, we have to determine the number of terms in the list 105, 112, 119, .... 994.
Clearly, it form an AP with first term a = 105 and common difference d = 112 – 105 = 7
Let there be n terms in the AP then, nth term = 994.
As ${a_n} = a + \left( {n - 1} \right)d$
$\Rightarrow$ 105 + (n – 1)7 = 994
$\Rightarrow$ 7(n – 1) = 994 – 105
$\Rightarrow$ 7 (n – 1) = 889
$\Rightarrow$ $n - 1 = {{889} \over 7} = 127$
$\Rightarrow$ n = 127 + 1 = 128
Hence, there are 128 numbers of three digit which are divisible by 7.

Q.14      How many multiples of 4 lie between 10 are 250 ?
Sol.         We observe that 12 is the first integer between 10 and 250, which is a multiple of 4 (i.e., divisible by 4). Also,   when we divide 250 by 4, the remainder is 2.
Therefore, 250 – 2 = 248 is the largest integer divisible by 4 (i.e., multiple of 4) and lying between 10 and 250. Thus, we have to find the number of terms in an AP with first term = 12, last term = 248, and common difference = 4 (as the numbers are divisible by 4).
Let there be n terms in the AP then.
${a_n} = 248$
$\Rightarrow$ 12 + (n – 1)4 = 248
$\Rightarrow$ 4(n – 1) = 248 – 12 $\Rightarrow$ 4(n – 1) = 236
$\Rightarrow$ $n - 1 = {{236} \over 4} = 59$
$\Rightarrow$ n = 59 + 1 = 60
Hence , there are 60 multiples of 4 between 10 and 250.

Q.15      For what value of n, are the nth terms of the APs : 63, 65, 67, .... and 3, 10, 17,... are equal?
Sol.             If nth terms of the APs 63, 65, 67, .... and 3, 10, 17, ..... are equal. Then,
63 + (n – 1) 2 = 3 + (n – 1) 7
[Since In 1st AP,  a = 63, d = 65 – 63 = 2 and in 2nd AP , a = 3, d = 10 – 3 = 7]
$\Rightarrow$ 7(n – 1) – 2 (n – 1) = 63 – 3
$\Rightarrow$ (n – 1) (7 – 2) = 60
$\Rightarrow$ 5(n – 1) = 60
$\Rightarrow$ $n - 1 = {{60} \over 5} = 12$
$\Rightarrow$ n = 12 + 1 = 13
Hence, the 13th terms of the two given APs are equal.

Q.16      Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Sol.           Let a be the first term and d the common difference.
Hence, ${a_3} = 16\,\,and\,{a_7} - {a_5} = 12$
$\Rightarrow$ a + 2d = 16 ... (1)
and, (a + 6d) – (a + 4d) = 12 $\Rightarrow$ 2d = 12
$\Rightarrow$ d = 6 ... (2)
Using (2) in (1), we get
a + 2 × 6 = 16
$\Rightarrow$ a = 16 – 12 = 4
Thus, the required AP is 4,  4 + 6,  4 + 2 × 6,  4 + 3 × 6,  ... i.e.,  4,  10,  16,  22, ....

Q.17      Find the 20th term from the last term of the AP : 3, 8, 13, .... 253.
Sol.          We have , l = Last term = 253
and d = Common ifference = 8 – 3 = 5
Therefore, 20th term from the end = l = (20 – 1)d
= l – 19 d
= 253 – 19 × 5
= 253 – 95 = 158

Q.18      The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
Sol.            Let a be the first term and d the common difference.
Here, ${a_4} + {a_8} = 24$
$\Rightarrow$ (a + 3d) + (a + 7d) = 24
$\Rightarrow$ 2a + 10d = 24
${a_6} + {a_10} = 44$
$\Rightarrow$ (a + 5d) + (a + 9d) = 44
$\Rightarrow$ 2a + 14d = 44 $\Rightarrow$ a + 7d = 22 ... (2)
Substracting (1) from (2), we get
2d = 10 $\Rightarrow$ d = 5
and then from (1)
a + 25 = 12 $\Rightarrow$ a = – 13
The first three terms are a, (a + d) and (a + 2d)
Putting values of a and d, we get – 13, (– 13 + 5) and (– 13 + 2 × 5)
i.e., – 13, – 8 and – 3

Q.19      Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?
Sol.           The annual salary drawn by Subha Rao in the years 1995, 1996, 1997, etc. is Rs 5,000, Rs 5,200, Rs 5,400.... Rs 7,000.
The list of these nos. is 5000, 5200, 5400, .... 7000.
It form an AP                               [since ${a_2} - {a_1} = {a_3} - {a_2} = 200$]
Let ${a_n} = 7000$
$\Rightarrow$ 7000 = a + (n – 1) d
$\Rightarrow$ 7000 = 5000 + (n – 1) (200)
$\Rightarrow$ 200 (n – 1) = 7000 – 5000
$\Rightarrow$ $n - 1 = {{2000} \over {200}} = 10$
$\Rightarrow$ n = 10 + 1 = 11
Thus, in the 11th year (i.e., in 2005) of his service, Subba Rao drew an annual salary of Rs. 7,000.

Q.20       Ramkali saves Rs 5 in the first week of a year and then increased her weekly savings by Rs 1.75. If in the nth week, her weekly savings become Rs 20.75, find n.
Sol.             Ramkali's savings in the subsequent weeks are respectively Rs 5, Rs 5 + Rs 1.75, Rs 5 + 2 × Rs 1.75, Rs 5 + 3 × Rs 1.75...
And in the nth week her saving will be Rs 5 + (n – 1) × Rs 1.75
$\Rightarrow$ 5 + (n – 1) × 1.75 = 20.75
$\Rightarrow$ (n – 1) × 1.75 = 20.75 – 5 = 15.75
Therefore $n - 1 = {{15.75} \over {1.75}} = 9$
$\Rightarrow$ n = 9 + 1 = 10

• Great job best of luck

• Wow great Sir great job

• satyarup

Thats a good app of complete our homework

• Nikhil

I am very happy with this i with this

• Armaanjit

It's very helpful for me. It helps in Mathematics

• THANKU

• very nice

• loveesh

very nice . thanks alot . i am very happy with this

• gourav dhiman

Very nice

• Anonymous

• Pratik

Nice one

• ASHISH BANSAL

VERY NICE YOU ALL ARE RIGHT

• Vansh

Awesome thanks allot dronstudy

• Vansh

Nice site have enjoyed

• Rohit

nice.................................................................................................................................................................
..............is very nic.............................................................................................

• diksh

its very nice

• diksh

its amazing

• Anonymous

its amazing

• rahul rajput

thank you very much for helping me

• Pranav.m

Thank you for help

• abhija

• srushti ganar

Thanks to dronstudy.com site.....

• JAYESH Babu

It is so interesting. I am very happy with this thanks a lot

• nandini

Iam happy with this

• Ronit kumar rohit

Thanks to dornstudy. .....
So amazing thinking. ....
Thanks a lot. ...