10th Maths 5.2

Chapter 5 Arithmetic Progressions

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NCERT Class 10th solution of Exercise 5.1

NCERT Class 10th solution of Exercise 5.3

NCERT Class 10th Maths Projects

Exercise 5.2

Q1. 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 ;

	$a$	$d$	$n$	$a_n$
i)	$7$	$3$	$8$	......
ii)	$-18$	.....	$10$	$0$
iii)	.....	$-3$	$18$	$-5$
iv)	$-18.9$	$2.5$	.....	$3.6$
v)	$3.5$	$0$	$105$	.....

Sol. :

$a_n=a+(n-1)timesd$

$a_n=7+(8-1)times3$

$a_n=7+7times3$

$a_n=7+21$

$a_n=28$

Answer :

$a_n=28$

ii)

$a_n=a+(n-1)timesd$

$0=-18+(10-1)timesd$

$0=-18+9timesd$

$9d=18$

$d=2$

Answer :

$d=2$ .

iii)

$a_n=a+(n-1)timesd$

$-5=a+(18-1)(-3)$

$-5=a+17(-3)$

$-5=a-51$

$a=46$

Answer :

$a=46$ .

iv)

$a_n=a+(n-1)timesd$

$3.6=-18.9+(n-1)times2.5$

$3.6=-18.9+2.5n-2.5$

$3.6=-21.4+2.5n$

$2.5n=-21.4-3.6$

$n=-25.0/2.5$

$n=-10$

Answer :

$n=-10$ .

$a_n=a+(n-1)timesd$

$a_n=3.5+(105-1)times0$

$a_n=3.5+0$

$a_n=3.5$

Answer :

$a_n=3.5$ .

Q2. Choose the correct choice is the following and justify :

i) $30^(th)$ term of the AP : $10, 7, 4,$ ...is

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

ii) $11^(th)$ term of the AP : $-3, -1/2, 2$ ,....is :

A) $28$ B) $22$ C) $-38$ D) $-48(1)/2$

Sol. :

i) Given :

$a=10, d=-3$ , and

$n=30$

$a_n=a+(n-1)d$

$a_(10)=10+(30-1)(-3)$

$a_(10)=10+29(-3)$

$a_(10)=10-87$

$a_(10)=-77$

Answer :

(C).

ii) Given :

$a=-3, d=2(1)/2$ , and

$n=11$

$a_n=a+(n-1)d$

$a_(11)=-3+(11-1)(5/2)$

$a_(11)=-3+25$

$a_(11)=22$

Answer :

(B).

Q3. In the following APs, find the missing terms in the boxes :

i) $2, square, 26$

ii) $square, 13, square, 3$

iii) $5, square, square, 9(1)/2$

iv) $-4, square, square, square, square, 6$

v) $square, 38, square, square, square, -22$

Sol. :

i) Given :

$a_1=2, a_3=26$ , and

$n=1, 3$

$a_n=a+(n-1)d$

$26=2+(3-1)d$

$26=2+2d$

$26-2=2d$

$24=2d$

$d=(24)/2$

$d=12$

$a_2=2+(2-1)12$

$a_2=2+12$

$a_2=14$

Answer :

$d=12, square_2=12$

ii) Given :

$a_2=13, a_4=3$ , and

$n=2, 4$

$a_n=a+(n-1)d$

$13=a+(2-1)d$

$13=a+d$ _________(1)

$3=a+(4-1)d$

$3=a+3d$ ___________(2)

By equation (1) - (2)

$13-3=a+d-(a+3d)$

$10=a+d-a-3d$

$10=(-2)d$

$d=(-10)/2$

$d=-5$

Put in (1)

$a+(-5)=13$

$a=18$

$a_3=18+(3-1)(-5)$

$a_3=18+2(-5)$

$a_3=18-10$

$a_3=8$

Answer :

$square_1=18, square_3=8$ .

iii) Given :

$a_1=5, a_4=9(1)/2=9.5$ , and

$n=4$

$a_n=a+(n-1)d$

$9.5=5+(4-1)d$

$9.5-5=3d$

$4.5=5d$

$d=(4.5)/5$

$d=1.5$

$a_2=a+(n-1)d$

$a_2=5+(2-1)1.5=6.5$

$a_3=a+(n-1)d$

$a_3=5+(3-1)1.5=8$

Answer :

$square_2=6.5, square_3=8$

iv) Given :

$a=-4, a_6=6$ , and

$n=6$

$a_n=a+(n-1)d$

$a_6=-4+(6-1)d$

$6=-4+5d$

$6+4=5d$

$10=5d$

$d=10/5$

$d=2$

$a_2=-4+(2-1)2=-4+2=-2$

$a_3=-4+(3-1)2=-4+4=0$

$a_4=-4+(4-1)2=-4+6=2$

$a_5=-4+(5-1)2=-4+8=4$

Answer :

$square_2=-2, square_3=0, square_4=2, and square_5=4$ .

v) Given :

$a_2=38, a_6=-22$ , and

$n=2, 6$

$a_n=a+(n-1)d$

$38=a+(2-1)d$

$38=a+d$ _________(1)

$-22=a+(6-1)d$

$-22=a+5d$ ________(2)

By equation (2) - (1)

$-22-38=a+5d-(a+d)$

$-60=a+5d-a-d$

$-60=4d$

$d=(-60)/4=-15$

Put in equation (1)

$38=a+(-15)$

$38=a-15$

$a=38+15=53$

$a_3=53+(3-1)(-15)=53-30=23$

$a_4=53+(4-1)(-15)=53-45=8$

$a_5=53+(5-1)(-15)=53-60=-7$

Answer :

$square_1=53, square_3=23, square_4=8$ , and

$square_5=-7$ .

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

Sol.

Given:

$a=3, d=8-3=5$ , and

$a_n=78$

$a_n=a+(n-1)d$

$78=3+(n-1)(5)$

$78=3+5n-5$

$78=-2+5n$

$5n=78+2$

$5n=80$

$n=80/5$

$n=16$

Answer :

$n=16$ .

Q5. Find the number of terms in each of the following APs :

i) $7, 13, 19, ...,205$ ii) $18, 15(1)/5, 13, .... -47$

Sol. :

i) Give:

$a=7, d=13-7=6$ , and

$a_n=205$

$a_n=a+(n-1)d$

$205=7+(n-1)(6)$

$205=7+6n-6$

$205=1+6n$

$6n=205-1$

$6n=204$

$n=204/6$

$n=34$

Answer :

$n=34$ .

ii) Given:

$a=18, d=15(1)/2-1=31/2-1=15.5-81=-2.5$ , and

$a_n=-47$

$a_n=a+(n-1)d$

$-47=18+(n-1)(-2.5)$

$-47=18-2.5n+2.5$

$-47=20.5-2.5n$

$2.5n=47+20.5$

$2.5n=67.5$

$n=(67.5)/(2.5)$

$n=(675)/25$

$n=27$

Answer :

$n=27$ .

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

Sol. : Given:

$a=11, d=8-11=-3$ , and

$a_n=-150$

$a_n=a+(n-1)d$

$-150=11+(n-1)(-3)$

$-150=11-3n+3$

$-150=14-3n$

$3n=14+150$

$3n=164$

$n=164/3$

$n=54(2)/3$ [ it is not natural number ]

Answer :

No,

$-150$ is not a term of given AP.

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

Sol. :

Given:

$a_11=38, a_16=73$ , and

$n=11, 16$

$a_n=a+(n-1)d$

$38=a+(11-1)d$

$38=a+10d$ ___________(1)

$73=a+(16-1)d$

$73=a+15d$ ___________(2)

By equation (2) - (1)

$73-38=a+15d-(a+10d)$

$35=a+15d-a-10d$

$35=5d$

$d=35/5$

$d=7$

Put in equation (1)

$38=a+10(7)$

$38=a+70$

$a=-70+38$

$a=-32$

Now

$a_31=a+(30-1)d$

$a_31=-32+30(7)$

$a_31=-32+210$

$a_31=178$

Answer :

$a_31=178$ .

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.

Sol. :

Given:

$a_3=12, a_50=106$ and

$n=12, 50$

$a_n=a+(n-1)d$

$12=a+(3-1)d$

$12=a+2d$ ________________(1)

$106=a+(50-1)d$

$106=a+49d$ ______________(2)

By equation (2) - (1)

$106-12=a+49d-(a+2d)$

$94=a+49d-a-2d$

$94=49d-2d$

$94=47d$

$d=94/47$

$d=2$

Put in equation (1)

$12=a+2(2)$

$12=a+4$

$a=8$

$a_29=a+(29-1)d$

$a_29=8+28(2)$

$a_29=8+56$

$a_29=64$

Answer :

$a_29=64$ .

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.

Sol. :

Given:

$a_3=4, a_9=-8$ , and

$a_n=0$

$a_n=a+(n-1)d$

$4=a+(3-1)d$

$4=a+2d$ ______________(1)

$-8=a+(9-1)d$

$-8=a+8d$ ______________(2)

By equation (2) - (1)

$-8-4=a+8d-(a+2d)$

$-12=a+8d-a-2d$

$-12=6d$

$d=-12/6$

$d=-2$

Put in equation (1)

$4=a+2(-2)$

$4=a-4$

$a=8$

Now

$a_n=8+(n-1)(-2)$

$0=8-2n+2$

$0=10-2n$

$2n=10$

$n=10/2$

$n=5$

Answer :

$n=5$ .

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

Sol. :

According to question

$a_17-a_10=7$

$a+(17-1)d-(a+(10-1)d=7$

$a+16d-a-9d=7$

$7d=7$

$d=7/7$

$d=1$

Answer :

$d=1$ .

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

Sol. :

Given:

$a=3, d=15-3=12$

According to question

$a_n-a_54=132$

$a+(n-1)d-[a+(54-1)d]=132$

$3+(n-1)12-[3+53(12)]=132$

$3+12n-12-3-636=132$

$12n=132+636+12$

$12n=780$

$n=780/12$

$n=65$

Answer :

$65$ th term.

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

Sol. :

Let first term of first AP is

$a$ ,

and first term of second AP is

$b$ .

According to question

$a_100-b_100=100$

$a+(100-1)d-[b+(100-1)d]=100$

$a+99d-b-99d=100$

`a-b=100___________________(1)

Now

$a_1000-b_1000$

$=a+(1000-1)d-[b+(1000-1)d]$

$=a+999d-b-999d$

$=a-b$

from equation (1)

$a-b=100$

Answer :

The difference between

$1000$ th terms will also be

$100$ .

Q13. How many three-digit numbers are divisible by $7$ ?

Sol. :

AP is

$105, 112, 119,.......994$ .

where

$a=105, d=7,$ and

$a_n=994$

$a_n=a+(n-1)d$

$994=105+(n-1)7$

$994=105+7n-7$

$994=98+7n$

$7n=994-98$

$7n=896$

$n=896/7$

$n=128$

Answer :

$n=128$ .

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

Sol. :

AP is

$12, 16, 20, 24, .....248$

where

$a=12, d=4,$ and

$a_n=248$

$a_n=a+(n-1)d$

$248=12+(n-1)4$

$248=12+4n-4$

$248=8+4n$

$4n=248-8$

$4n=240$

$n=240/4$

$n=60$

Answer :

$n=60$ .

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

Sol. :

AP `63, 65, 67,.....

where

$a=63, d=65-63=2$

AP `3, 10, 17, ....

where

$a=3, d=10-3=7$

According to question

$a_n=b_n$

$63+(n-1)2=3+(n-1)7$

$63+2n-2=3+7n-7$

$7n-2n=63-3-2+7$

$5n=65$

$n=65/5$

$n=13$

Answer :

$n=13$ .

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

Sol. :

Let the AP is

$a, a+d, a+2d, .....$

According to question

$a_3=a+2d$

$16=a+2d$ ________(1)

and

$a_7-a_5=12$

$a+(7-1)d-[a+(5-1)d]=12$

$a+6d-a-4d=12$

$2d=12$

$d=6$

Put in equation (1)

$16=a+2(6)$

$16=a+12$

$a=4$

$a_2=4+(2-1)6=4+6=10$

$a_3=4+(3-1)6=4+12=16$

$a_4=4+(4-1)6=4+18=22$

Answer :

The AP is

$4, 10, 16, 22, .....$ .

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

Sol. :

By the given AP take in decending order

$253, 248, ......, 8, 3$

where

$a=253, d=248-253=-5$

$a_20=253+(20-1)(-5)$

$a_20=253+19(-5)$

$a_20=253-95$

$a_20=158$

Answer :

$a_20=158$ .

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 term is $44$ . Find the first three terms of the AP.

Sol. :

Let the AP is

$a, a+d, a+2d, ......$

According to question

$a_4+a_8=24$

$a+(4-1)d+a+(8-1)d=24$

$a+3d+a+7d=24$

$2a+10d=24$

$a+5d=12$ ______________(1)

and

$a_6+a_10=44$

$a+(6-1)d+a+(10-1)d=44$

$a+5d+a+9d=44$

$2a+14d=44$

$a+7d=22$ _____________(2)

By equation (2) - (1)

$a+7d-(a+5d)=22-12$

$a+7d-a-5d=10$

$2d=10$

$d=10/2$

$d=5$

Put in equation (1)

$a+5(5)=12$

$a+25=12$

$a=-13$

$a_2=-13+(2-1)5=-13+5=-8$

$a_3=-13+(3-1)5=-13+10=-3$

Answer :

The first three terms of AP is

$-13, -8, -3, ....$ .

Q19. Subba Rao started work in $1995$ at an annual salary of $₹5000$ and received an increment of $₹ 200$ each year. In which year did his income reach $₹ 7000$ ?

Sol. :

Given:

$a=₹ 5000, d=₹ 200,$ and

$a_n=₹ 7000$ .

$a_n=a+(n-1)d$

$7000=5000+(n-1)200$

$7000=5000+200n-200$

$200n=7000-5000+200$

`n=2200/200

$n=11$

Answer :

$11$ th year.

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

Sol. :

Given:

$a=₹ 5, d=₹ 1.75$ and

$a_n=₹ 20.75$

$a_n=a+(n-1)d$

$20.75=5+(n-1)1.75$

$20.75=5+1.75n-1.75$

$1.75n=20.75-5+1.75$

$n=17.50/1.75$

$n=10$

Answer :

$10$ th year.

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10th Maths 5.2

Chapter 5

Arithmetic Progressions

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NCERT Class 10th solution of Exercise 5.1

NCERT Class 10th solution of Exercise 5.3

NCERT Class 10th Maths Projects

Exercise 5.2

$a$

$d$

$n$

$a_n$

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NCERT Class 10th solution of Exercise 5.1

NCERT Class 10th solution of Exercise 5.3

NCERT Class 10th Maths Projects

Exercise 5.2

aa

dd

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ana_n

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10th Maths Chapter 1

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$a$

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