NCERT Solutions for Exercise 5.2 Class 10 Maths Chapter 5 - Arithmetic Progressions

Answer:
(i) Given $a=7,d=3,n=8$
Now, we know that
$a_n=a+(n-1)d$
$a_8=7+(8-1)3=7+7\times 3=7+21=28$
Therefore,
$a_8=28$

(ii) Given $a=-18,n=10,a_{10}=0$
Now, we know that
$a_n=a+(n-1)d$
$a_{10}=-18+(10-1)d$
$0+18=9d$
$d=\frac{18}{9}=2$

(iii) Given $d=-3,n=18,a_{18}=-5$
Now, we know that
$a_n=a+(n-1)d$
$a_{18}=a+(18-1)(-3)$
$-5=a+17\times (-3)$
$a=51-5=46$
Therefore,
$a=46$

(iv) Given $a=-18.9,d=2.5,a_n=3.6$
Now, we know that
$a_n=a+(n-1)d$
$a_n=-18.9+(n-1)2.5$
$3.6+18.9=2.5n-2.5$
$n=\frac{22.5+2.5}{2.5}=\frac{25}{2.5}=10$
Therefore,
$n=10$

(v) Given $a=3.5,d=0,n=105$
Now, we know that
$a_n=a+(n-1)d$
$a_{105}=3.5+(105-1)0$
$a_{105}=3.5$
Therefore,
$a_{105}=3.5$

Q2 (i) Choose the correct choice in the following and justify: $30$ th term of the AP: $10,7,4,...,$ is

(A) $97$ (B) $77$ (C) $-77$ (D) $-87$

Answer:
Given series $10,7,4,...,$
Here, $a=10$ and $d=7-10=-3$
Now, we know that
$a_n=a+(n-1)d$
It is given that $n=30$
Therefore, after putting values, we get:
$a_{30}=10+(30-1)(-3)$
$a_{30}=10+(29)(-3)$
$a_{30}=10-87=-77$
Therefore, $30$ th term of the AP: $10,7,4,...,$ is -77
Hence, the correct answer is (C)

Q2 (ii) Choose the correct choice in the following and justify : 11th term of the AP: $-3,-\frac{1}{2},2,...,$ is

(A) $28$ (B) $22$ (C) $-38$ (D) $-48\frac{1}{2}$

Answer:
Given series $-3,-\frac{1}{2},2,...,$
Here, $a=-3$ and $d=-\frac{1}{2}-(-3)=-\frac{1}{2}+3=\frac{-1+6}{2}=\frac{5}{2}$
Now, we know that
$a_n=a+(n-1)d$
It is given that $n=11$
Therefore, after putting values, we get:
$a_{11}=-3+(11-1)\left(\frac{5}{2}\right)$
$a_{11}=-3+(10)\left(\frac{5}{2}\right)$
$a_{11}=-3+5\times 5=-3+25=22$
Therefore, 11th term of the AP: $-3,-\frac{1}{2},2,...,$ is 22
Hence, the Correct answer is (B)

Q3 (i) In the following APs, find the missing terms in the boxes: $2,\boxed{},26$

Answer:
Given series $2,\boxed{},26$
Here, $a=2,n=3and{a}_3=26$
Now, we know that
$a_n=a+(n-1)d$
$\Rightarrow a_3=2+(3-1)d$
$\Rightarrow 26-2=(2)d$
$\Rightarrow d=\frac{24}{2}=12$
Now, after putting values we get:
$a_2=a_1+d$
$a_2=2+12=14$
Therefore, the missing term is 14

Q3 (ii) In the following APs, find the missing terms in the boxes: $\boxed{},13,\boxed{},3$

Answer:
Given AP series is
$\boxed{},13,\boxed{},3$
Here, $a_2=13,n=4and{a}_4=3$
Now,
$a_2=a_1+d$
$a_1=a=13-d$
Now, we know that
$a_n=a+(n-1)d$
$\Rightarrow a_4=13-d+(4-1)d$
$\Rightarrow 3-13=-d+3d$
$\Rightarrow d=-\frac{10}{2}=-5$
Now, after putting values we get:
$a_2=a_1+d$
$a_1=a=13-d=13-(-5)=18$
And
$a_3=a_2+d$
$a_3=13-5=8$
Therefore, missing terms are 18 and 8
AP series is 18,13,8,3

Q3 (iii) In the following APs, find the missing terms in the boxes: $5,\boxed{},\boxed{},9\frac{1}{2}$

Answer:
Given AP series is $5,\boxed{},\boxed{},9\frac{1}{2}$
Here, $a=5,n=4and{a}_4=9\frac{1}{2}=\frac{19}{2}$
Now, we know that
$a_n=a+(n-1)d$
$\Rightarrow a_4=5+(4-1)d$
$\Rightarrow \frac{19}{2}-5=3d$
$\Rightarrow d=\frac{19-10}{2\times 3}=\frac{9}{6}=\frac{3}{2}$
Now, after putting values we get:
$a_2=a_1+d$
$a_2=5+\frac{3}{2}=\frac{13}{2}$
And
$a_3=a_2+d$
$a_3=\frac{13}{2}+\frac{3}{2}=\frac{16}{2}=8$
Therefore, missing terms are $\frac{13}{2}$ and 8
AP series is $5,\frac{13}{2},8,\frac{19}{2}$

Q3 (iv) In the following APs, find the missing terms in the boxes: $-4,\boxed{},\boxed{},\boxed{},\boxed{},6$
Answer:

Given AP series is $-4,\boxed{},\boxed{},\boxed{},\boxed{},6$
Here, $a=-4,n=6and{a}_6=6$
Now, we know that
$a_n=a+(n-1)d$
$\Rightarrow a_6=-4+(6-1)d$
$\Rightarrow 6+4=5d$
$\Rightarrow d=\frac{10}{5}=2$
Now, after putting values we get:
$a_2=a_1+d$
$a_2=-4+2=-2$
And
$a_3=a_2+d$
$a_3=-2+2=0$
And
$a_4=a_3+d$
$a_4=0+2=2$
And
$a_5=a_4+d$
$a_5=2+2=4$
Therefore, missing terms are -2, 0, 2, 4
AP series is -4, -2, 0, 2, 4, 6

Q3 (v) In the following APs, find the missing terms in the boxes: $\boxed{},38,\boxed{},\boxed{},\boxed{},-22$

Answer:
Given the AP series is
$\boxed{},38,\boxed{},\boxed{},\boxed{},-22$
Here, $a_2=38,n=6and{a}_6=-22$
Now,
$a_2=a_1+d$
$a_1=a=38-d-(i)$
Now, we know that
$a_n=a+(n-1)d$
$\Rightarrow a_6=38-d+(6-1)d(using(i))$
$\Rightarrow -22-38-=-d+5d$
$\Rightarrow d=-\frac{60}{4}=-15$
Now, after putting values we get:
$a_2=a_1+d$
$a_1=38-(-15)=38+15=53$
And
$a_3=a_2+d$
$a_3=38-15=23$
And
$a_4=a_3+d$
$a_4=23-15=8$
And
$a_5=a_4+d$
$a_5=8-15=-7$
Therefore, missing terms are 53, 23, 8, -7
AP series is 53, 38, 23, 8, -7, -22

Q4 Which term of the AP : $3,8,13,18,...,$ is $78$ ?

Answer:
Given AP series $3,8,13,18,...,$
Let suppose that nth term of AP is 78
Here, $a=3$ and $d=a_2-a_1=8-3=5$
Now, we know that that
$a_n=a+(n-1)d$
$\Rightarrow 78=3+(n-1)5$
$\Rightarrow 78-3=5n-5$
$\Rightarrow n=\frac{75+5}{5}=\frac{80}{5}=16$
Therefore, value of 16th term of given AP is 78

Q5 (i) Find the number of terms in each of the following APs: $7,13,19,...,205$

Answer:
Given AP series $7,13,19,...,205$
Let's suppose there are n terms in given AP
Then,
$a=7,a_n=205$
And
$d=a_2-a_1=13-7=6$
Now, we know that
$a_n=a+(n-1)d$
$\Rightarrow 205=7+(n-1)6$
$\Rightarrow 205-7=6n-6$
$\Rightarrow n=\frac{198+6}{6}=\frac{204}{6}=34$
Therefore, there are 34 terms in given AP

Q5 (ii) Find the number of terms in each of the following APs: $18,15\frac{1}{2},13,...,-47$

Answer:
Given AP series $18,15\frac{1}{2},13,...,-47$
suppose there are n terms in given AP
Then,
$a=18,a_n=-47$
And
$d=a_2-a_1=\frac{31}{2}-18=\frac{31-36}{2}=-\frac{5}{2}$
Now, we know that
$a_n=a+(n-1)d$
$\Rightarrow -47=18+(n-1)(-\frac{5}{2})$
$\Rightarrow -47-18=-\frac{5n}{2}+\frac{5}{2}$
$\Rightarrow -\frac{5n}{2}=-65-\frac{5}{2}$
$\Rightarrow -\frac{5n}{2}=-\frac{135}{2}$
$\Rightarrow n=27$
Therefore, there are 27 terms in given AP

Q6 Check whether $-150$ is a term of the AP : $11,8,5,2...$

Answer:
Given AP series $11,8,5,2...$
Here, $a=11$ and $d=a_2-a_1=8-11=-3$
Now,
suppose -150 is nth term of the given AP
Now, we know that
$a_n=a+(n-1)d$
$\Rightarrow -150=11+(n-1)(-3)$
$\Rightarrow -150-11=-3n+3$
$\Rightarrow =n=\frac{161+3}{3}=\frac{164}{3}=54.66$
Value of n is not an integer
Therefore, -150 is not a term of AP $11,8,5,2...$

Q7 Find the $31$ st term of an AP whose $11$ th term is $38$ and the $16$ th term is $73$ .

Answer:
Given: $11$ th term of an AP is $38$ and the $16$ th term is $73$
Now,
$a_{11}=38=a+10d-(i)$
And
$a_{16}=73=a+15d-(ii)$
On solving equation (i) and (ii) we will get
$a=-32andd=7$
Now,
$a_{31}=a+30d=-32+30\times 7=-32+210=178$
Therefore, 31st terms of given AP is 178

Q8 An AP consists of $50$ terms of which $3$ rd term is $12$ and the last term is $106$ . Find the $29$ th term.
Answer:

Given: AP consists of $50$ terms of which $3$ rd term is $12$ and the last term is $106$
Now,
$a_3=12=a+2d-(i)$
And
$a_{50}=106=a+49d-(ii)$
On solving equation (i) and (ii) we will get
$a=8andd=2$
Now,
$a_{29}=a+28d=8+28\times 2=8+56=64$
Therefore, 29th term of given AP is 64

Q9 If the $3$ rd and the $9$ th terms of an AP are $4$ and $-8$ respectively, which term of this AP is zero?
Answer:

Given: $3$ rd and the $9$ th terms of an AP are $4$ and $-8$ respectively
Now,
$a_3=4=a+2d-(i)$
And
$a_9=-8=a+8d-(ii)$
On solving equation (i) and (ii) we will get
$a=8andd=-2$
Now,
Let nth term of given AP is 0
Then,
$a_n=a+(n-1)d$
$0=8+(n-1)(-2)$
$2n=8+2=10$
$n=\frac{10}{2}=5$
Therefore, 5th term of given AP is 0

Q10 The $17$ th term of an AP exceeds its $10$ th term by $7$ . Find the common difference.
Answer:

Given: $17$ th term of an AP exceeds its $10$ th term by $7$
i.e.
$a_{17}=a_{10}+7$
$\Rightarrow a+16d=a+9d+7$
$\Rightarrow a+16d-a-9d=7$
$\Rightarrow 7d=7$
$\Rightarrow d=1$
Therefore, the common difference of AP is 1

Q11 Which term of the AP : $3,15,27,39,...$ will be $132$ more than its $54$ th term?
Answer:

Given series $3,15,27,39,...$
Here, $a=3$ and $d=a_2-a_1=15-3=12$
Now, let's suppose nth term of given AP is $132$ more than its $54$ th term
Then,
$a_n=a_{54}+132$
$\Rightarrow a+(n-1)d=a+53d+132$
$\Rightarrow 3+(n-1)12=3+53\times 12+132$
$\Rightarrow 12n=3+636+132+12$
$\Rightarrow 12n=636+132+12$
$\Rightarrow n=\frac{780}{12}=65$
Therefore, 65th term of given AP is $132$ more than its $54$ th term

Q12 Two APs have the same common difference. The difference between their $100$ th terms is $100$ , what is the difference between their $1000$ th terms?
Answer:

Given: Two APs have the same common difference and difference between their $100$ th terms is $100$
i.e.
$a_{100}-a_{100}^{\prime }=100$
Let common difference of both the AP's is d
$\Rightarrow a+99d-a^{\prime }-99d=100$
$\Rightarrow a-a^{\prime }=100-(i)$
Now, difference between 1000th term is
$a_{1000}-a_{1000}^{\prime }$
$\Rightarrow a+999d-a^{\prime }-999d$
$\Rightarrow a-a^{\prime }$
$\Rightarrow 100(using(i))$
Therefore, difference between 1000th term is 100

Q 13 How many three-digit numbers are divisible by $7$ ?

Answer:

We know that the first three-digit number divisible by 7 is 105 and last three-digit number divisible by 7 is 994
Therefore,
$a=105,d=7and{a}_n=994$
Let there are n three digit numbers divisible by 7
Now, we know that
$a_n=a+(n-1)d$
$\Rightarrow 994=105+(n-1)7$
$\Rightarrow 7n=896$
$\Rightarrow n=\frac{896}{7}=128$
Therefore, there are 128 three-digit numbers divisible by 7

Q14 How many multiples of $4$ lie between $10$ and $250$ ?
Answer:

We know that the first number divisible by 4 between 10 to 250 is 12 and last number divisible by 4 is 248
Therefore,
$a=12,d=4and{a}_n=248$
Let there are n numbers divisible by 4
Now, we know that
$a_n=a+(n-1)d$
$\Rightarrow 248=12+(n-1)4$
$\Rightarrow 4n=240$
$\Rightarrow n=\frac{240}{4}=60$
Therefore, there are 60 numbers between 10 to 250 that are divisible by 4

Q15 For what value of $n$ , are the $n$ th terms of two APs: $63,65,67,...$ and $3,10,17,...$ equal?

Answer:
Given two AP's are
$63,65,67,...$ and $3,10,17,...$
Let first term and the common difference of two AP's are a , a' and d , d'
$a=63,d=a_2-a_1=65-63=2$
And
$a^{\prime }=3,d^{\prime }=a_2^{\prime }-a_1^{\prime }=10-3=7$
Now,
Let nth term of both the AP's are equal
$a_n=a_n^{\prime }$
$\Rightarrow a+(n-1)d=a^{\prime }+(n-1)d^{\prime }$
$\Rightarrow 63+(n-1)2=3+(n-1)7$
$\Rightarrow 5n=65$
$\Rightarrow n=\frac{65}{5}=13$
Therefore, the 13th term of both the AP's are equal

Q16 Determine the AP whose third term is $16$ and the $7$ th term exceeds the $5$ th term by $12$ .
Answer:

It is given that
3rd term of AP is $16$ and the $7$ th term exceeds the $5$ th term by $12$
i.e.
$a_3=a+2d=16-(i)$
And
$a_7=a_5+12$
$a+6d=a+4d+12$
$2d=12$
$d=6$
Put the value of d in equation (i) we will get
$a=4$
Now, AP with first term = 4 and common difference = 6 is
4,10,16,22,.....

Q17 Find the $20$ th term from the last term of the AP : $3,8,13,...,253$ .

Answer:
Given AP is
$3,8,13,...,253$
Here, $a=3and{a}_n=253$
And
$d=a_2-a_1=8-3=5$
Let suppose there are n terms in the AP
Now, we know that
$a_n=a+(n-1)d$
$253=3+(n-1)5$
$5n=255$
$n=51$
So, there are 51 terms in the given AP and 20th term from the last will be 32th term from the starting
Therefore,
$a_{32}=a+31d$
$a_{32}=3+31\times 5=3+155=158$
Therefore, 20th term from the of given AP is 158

Q18 The sum of the $4$ th and $8$ th terms of an AP is $24$ and the sum of the $6$ th and $10$ th terms is $44$ Find the first three terms of the AP.
Answer:

It is given that
sum of the < img alt="\small 4" class="fr-fic fr-dii" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Csmall%204"> th and $8$ th terms of an AP is $24$ and the sum of the $6$ th and $10$ th terms is $44$
i.e.
$a_4+a_8=24$
$\Rightarrow a+3d+a+7d=24$
$\Rightarrow 2a+10d=24$
$\Rightarrow a+5d=12-(i)$
And
$a_6+a_{10}=44$
$\Rightarrow a+5d+a+9d=44$
$\Rightarrow 2a+14d=44$
$\Rightarrow a+7d=22-(ii)$
On solving equation (i) and (ii) we will get
$a=-13andd=5$
Therefore,first three of AP with a = -13 and d = 5 is
-13,-8,-3

Q19 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?

Answer:
It is given that
Subba Rao started work at an annual salary of Rs 5000 and received an increment of Rs 200 each year
Therefore, $a=5000andd=200$
Let's suppose after n years his salary will be Rs 7000
Now, we know that
$a_n=a+(n-1)d$
$\Rightarrow 7000=5000+(n-1)200$
$\Rightarrow 2000=200n-200$
$\Rightarrow 200n=2200$
$\Rightarrow n=11$
Therefore, after 11 years his salary will be Rs 7000
after 11 years, starting from 1995, his salary will reach to 7000, so we have to add 10 in 1995, because these numbers are in years
Thus , 1995+10 = 2005

Q20 Ramkali saved Rs 5 in the first week of a year and then increased her weekly savings by Rs $1.75$ . If in the $n$ th week, her weekly savings become Rs $20.75$ , find $n$

Answer:
It is given that
Ramkali saved Rs 5 in the first week of a year and then increased her weekly savings by Rs $1.75$
Therefore, $a=5andd=1.75$
after $n$ th week, her weekly savings become Rs $20.75$
Now, we know that
$a_n=a+(n-1)d$
$\Rightarrow 20.75=5+(n-1)1.75$
$\Rightarrow 15.75=1.75n-1.75$
$\Rightarrow 1.75n=17.5$
$\Rightarrow n=10$
Therefore, after 10 weeks, her savings will become Rs 20.75