Solution

I. $a=7, d=3, n=8, a_n=$ ?

We know that,

For an A.P. $a_n=a+(n-1) d$

$=7+(8-1) 3$

$=7+(7) 3$

$=7+21=28$

Hence, $a_n=28$

II. Given that

$a=-18, n=10, a_n=0, d=$ ?

We know that,

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

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

$18=9 d$

$d=\dfrac{18}{9}=2$

Hence, common difference, $d=2$

III. Given that

$d=-3, n=18, a_n=-5$

We know that,

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

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

$-5=a+(17)(-3)$ $-5=a-51$

$a=51-5=46$

Hence, $a=46$

IV. $a=-18.9, d=2.5, a_n=3.6, n=$ ?

We know that,

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

$3.6=-18.9+(n-1) 2.5$

$3.6+18.9=(n-1) 2.5$

$22.5=(n-1) 2.5$

$(n-1)=\dfrac{22.5}{2.5}$

$n-1=9$

$n=10$

Hence, $n=10$

V. $a=3.5, d=0, n=105, a_n=$ ?

We know that,

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

$a_n=3.5+(105-1) 0$

$a_n=3.5+104 \times 0$

$a_n=3.5$

Hence, $a_n=3.5$

Solution

I. Given that

A.P. $10,7,4, \ldots$

First term, $a=10$

Common difference, $d=a_2-a_1=7-10$

$=-3$

We know that, $a_n=a+(n-1) d$

$a _{30}=10+(30-1)(-3)$

$a _{30}=10+(29)(-3)$

$a _{30}=10-87=-77$

Hence, the correct answer is $\mathbf{C}$.

II. Given that, A.P.

$ -3,-\dfrac{1}{2}, 2, \ldots $

First term $a=-3$

Common difference, $d=a_2-a_1$

$=-\dfrac{1}{2}-(-3)$

$=-\dfrac{1}{2}+3=\dfrac{5}{2}$

We know that,

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

$a _{11}=-3+(11-1)(\dfrac{5}{2})$

$a _{11}=-3+(10)(\dfrac{5}{2})$

$a _{11}=-3+25$

$a _{11}=22$

Hence, the answer is $\mathbf{B}$.

Solution

I. $2, \square, 26$

For this A.P.,

$a=2$

$a_3=26$

We know that, $a_n=a+(n-1) d$

$a_3=2+(3-1) d$

$26=2+2 d$

$24=2 d$

$d=12$

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

$=14$

Therefore, 14 is the missing term.

II. $13, \square, 3$

For this A.P.,

$a_2=13$ and

$a_4=3$

We know that, $a_n=a+(n-1) d$

$a_2=a+(2-1) d$

$13=a+d(1)$

$a_4=a+(4-1) d$

$3=a+3 d$ (II)

On subtracting (I) from (II), we obtain

$-10=2 d$

$d=-5$

From equation (I), we obtain

$13=a+(-5)$

$a=18$

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

$=18+2(-5)=18-10=8$

Therefore, the missing terms are 18 and 8 respectively.

III.

$5, \square, \square, 9 \dfrac{1}{2}$

For this A.P.,

$a=5$

$a_4=9 \dfrac{1}{2}=\dfrac{19}{2}$

We know that,

$a_n=a+(n-1) d$ $a_4=a+(4-1) d$

$\dfrac{19}{2}=5+3 d$

$\dfrac{19}{2}-5=3 d$

$\dfrac{9}{2}=3 d$

$d=\dfrac{3}{2}$

$a_2=a+d=5+\dfrac{3}{2}=\dfrac{13}{2}$

$a_3=a+2 d=5+2(\dfrac{3}{2})=8$

Therefore, the missing terms are $\dfrac{13}{2}$ and 8 respectively.

IV.

-4 $\square$, $\square$, $\square$, $\square$, 6

For this A.P.,

$a=-4$ and

$a_6=6$

We know that,

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

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

$6=-4+5 d$

$10=5 d$

$d=2$

$a_2=a+d=-4+2=-2$ $a_3=a+2 d=-4+2(2)=0$

$a_4=a+3 d=-4+3(2)=2$

$a_5=a+4 d=-4+4(2)=4$

Therefore, the missing terms are $-2,0,2$, and 4 respectively.

v.

$\square$, 38, $\square$, $\square$, $\square$, -22

For this A.P.,

$a_2=38$

$a_6=-22$

We know that

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

$a_2=a+(2-1) d$

$38=a+d(1)$

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

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

On subtracting equation (1) from (2), we obtain

$-22-38=4 d$

$-60=4 d$

$d=-15$

$a=a_2-d=38-(-15)=53$

$a_3=a+2 d=53+2(-15)=23$

$a_4=a+3 d=53+3(-15)=8$

$a_5=a+4 d=53+4(-15)=-7$

Therefore, the missing terms are $53,23,8$, and -7 respectively.

Solution

$3,8,13,18, \ldots$

For this A.P.,

$a=3$

$d=a_2-a_1=8-3=5$

Let $n^{\text{th }}$ term of this A.P. be 78 . $a_n=a+(n-1) d$

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

$75=(n-1) 5$

$(n-1)=15$

$n=16$

Hence, $16^{\text{th }}$ term of this A.P. is 78.

Solution

I. $7,13,19, \ldots, 205$

For this A.P.,

$a=7$

$d=a_2-a_1=13-7=6$

Let there are $n$ terms in this A.P.

$a_n=205$

We know that

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

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

$198=(n-1) 6$

$33=(n-1)$

$n=34$

Therefore, this given series has 34 terms in it.

II.

$18,15 \dfrac{1}{2}, 13, \ldots,-47$

For this A.P., $a=18$

$d=a_2-a_1=15 \dfrac{1}{2}-18$

$d=\dfrac{31-36}{2}=-\dfrac{5}{2}$

Let there are $n$ terms in this A.P.

Therefore, $a_n=-47$ and we know that,

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

$-47=18+(n-1)(-\dfrac{5}{2})$

$-47-18=(n-1)(-\dfrac{5}{2})$

$-65=(n-1)(-\dfrac{5}{2})$

$(n-1)=\dfrac{-130}{-5}$

$(n-1)=26$

$n=27$

Therefore, this given A.P. has 27 terms in it.

Solution

For this A.P.,

$a=11$

$d=a_2-a_1=8-11=-3$

Let -150 be the $n^{\text{th }}$ term of this A.P.

We know that, $a_n=a+(n-1) d$

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

$-150=11-3 n+3$

$-164=-3 n$

$n=\dfrac{164}{3}$

Clearly, $n$ is not an integer.

Therefore, - 150 is not a term of this A.P.

Solution

Given that,

$a _{11}=38$

$a _{16}=73$

We know that,

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

$a _{11}=a+(11-1) d$

$38=a+10 d(1)$

Similarly,

$a _{16}=a+(16-1) d$

$73=a+15 d(2)$

On subtracting (1) from (2), we obtain

$35=5 d$

$d=7$

From equation (1),

$38=a+10 \times(7)$

$38-70=a$

$a=-32$

$a _{31}=a+(31-1) d$

$=-32+30(7)$

$=-32+210$

$=178$

Hence, $31^{\text{st }}$ term is 178.

Solution

Given that,

$a_3=12$

$a _{50}=106$

We know that,

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

$a_3=a+(3-1) d$

$12=a+2 d(1)$

Similarly, $a _{50}=a+(50-1) d$

$106=a+49 d$ (II)

On subtracting (I) from (II), we obtain

$94=47 d$

$d=2$

From equation (I), we obtain

$12=a+2(2)$

$a=12-4=8$

$a _{29}=a+(29-1) d$

$a _{29}=8+(28) 2$

$a _{29}=8+56=64$

Therefore, $29^{\text{th }}$ term is 64 .

Solution

Given that,

$a_3=4$

$a_9=-8$

We know that,

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

$a_3=a+(3-1) d$

$4=a+2 d(1)$

$a_9=a+(9-1) d$

$-8=a+8 d$ (II)

On subtracting equation (I) from (II), we obtain

$-12=6 d$

$d=-2$

From equation (I), we obtain

$4=a+2(-2)$

$4=a-4$

$a=8$

Let $n^{\text{th }}$ term of this A.P. be zero.

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

$0=8+(n-1)(-2)$

$0=8-2 n+2$

$2 n=10$

$n=5$

Hence, $5^{\text{th }}$ term of this A.P. is 0 .

Solution

We know that,

For an A.P., $a_n=a+(n-1) d$

$a _{17}=a+(17-1) d$

$a _{17}=a+16 d$

Similarly, $a _{10}=a+9 d$

It is given that

$a _{17}-a _{10}=7$

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

$7 d=7$ $d=1$

Therefore, the common difference is 1 .

Solution

Given A.P. is $3,15,27,39, \ldots$

$a=3$

$d=a_2-a_1=15-3=12$

$a _{54}=a+(54-1) d$

$=3+(53)(12)$

$=3+636=639$

$132+639=771$

We have to find the term of this A.P. which is 771 .

Let $n^{\text{th }}$ term be 771 .

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

$771=3+(n-1) 12$

$768=(n-1) 12$

$(n-1)=64$

$n=65$

Therefore, $65^{\text{th }}$ term was 132 more than $54^{\text{th }}$ term.

Alternatively,

Let $n^{\text{th }}$ term be 132 more than $54^{\text{th }}$ term.

$n=54+\dfrac{132}{12}$

$=54+11=65^{\text{th }}$ term

Solution

Let the first term of these A.P.s be $a_1$ and $a_2$ respectively and the common difference of these A.P.s be $d$.

For first A.P.,

$a _{100}=a_1+(100-1) d$

$=a_1+99 d$

$a _{1000}=a_1+(1000-1) d$

$a _{1000}=a_1+999 d$

For second A.P.,

$a _{100}=a_2+(100-1) d$

$=a_2+99 d$

$a _{1000}=a_2+(1000-1) d$

$=a_2+999 d$

Given that, difference between

$100^{\text{th }}$ term of these A.P.s $=100$

Therefore, $(a_1+99 d)-(a_2+99 d)=100$

$a_1-a_2=100$ (1)

Difference between $1000^{\text{th }}$ terms of these A.P.s

$(a_1+999 d)-(a_2+999 d)=a_1-a_2$

From equation (1),

This difference, $a_1-a_2=100$

Hence, the difference between $1000^{\text{th }}$ terms of these A.P. will be 100 .

Solution

First three-digit number that is divisible by $7=105$

Next number $=105+7=112$

Therefore, $105,112,119, \ldots$

All are three digit numbers which are divisible by 7 and thus, all these are terms of an A.P. having first term as 105 and common difference as 7 .

The maximum possible three-digit number is 999 . When we divide it by 7 , the remainder will be 5 . Clearly, $999-5=$ 994 is the maximum possible three-digit number that is divisible by 7 .

The series is as follows.

105, 112, 119, …, 994

Let 994 be the $n$th term of this A.P. $a=105$

$d=7$

$a_n=994$

$n=$ ?

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

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

$889=(n-1) 7$

$(n-1)=127$

$n=128$

Therefore, 128 three-digit numbers are divisible by 7 .

Solution

First multiple of 4 that is greater than 10 is 12 . Next will be 16 .

Therefore, 12, 16, 20, 24, …

All these are divisible by 4 and thus, all these are terms of an A.P. with first term as 12 and common difference as 4 .

When we divide 250 by 4 , the remainder will be 2 . Therefore, $250-2=248$ is divisible by 4 .

The series is as follows.

$12,16,20,24, \ldots, 248$

Let 248 be the $n^{\text{th }}$ term of this A.P.

$a=12$

$d=4$

$a_n=248$

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

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

$\dfrac{236}{4}=n-1$

$59=n-1$

$n=60$

Therefore, there are 60 multiples of 4 between 10 and 250 .

Solution

$63,65,67, \ldots$

$a=63$

$d=a_2-a_1=65-63=2$

$n^{\text{th }}$ term of this A.P. $=a_n=a+(n-1) d$

$a_n=63+(n-1) 2=63+2 n-2$

$a_n=61+2 n(1)$

$3,10,17, \ldots$

$a=3$

$d=a_2-a_1=10-3=7$

$n^{\text{th }}$ term of this A.P. $=3+(n-1) 7$

$a_n=3+7 n-7$

$a_n=7 n-4(2)$

It is given that, $n^{\text{th }}$ term of these A.P.s are equal to each other.

Equating both these equations, we obtain

$61+2 n=7 n-4$

$61+4=5 n$

$5 n=65$

$n=13$

Therefore, $13^{\text{th }}$ terms of both these A.P.s are equal to each other.

Solution

$=a_3=16$

$a+(3-1) d=16$

$a+2 d=16(1)$ $a_7-a_5=12$

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

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

$2 d=12$

$d=6$

From equation (1), we obtain

$a+2(6)=16$

$a+12=16$

$a=4$

Therefore, A.P. will be

$4,10,16,22, \ldots$

Solution

Given A.P. is

$3,8,13, \ldots, 253$

Common difference for this A.P. is 5 .

Therefore, this A.P. can be written in reverse order as

$253,248,243, \ldots, 13,8,5$

For this A.P.,

$a=253$

$d=248-253=-5$

$n=20$

$a _{20}=a+(20-1) d$

$a _{20}=253+(19)(-5)$

$a _{20}=253-95$

$a=158$

Therefore, $20^{\text{th }}$ term from the last term is 158.

Solution

We know that,

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

$a_4=a+(4-1) d$

$a_4=a+3 d$

Similarly,

$a_8=a+7 d$

$a_6=a+5 d$

$a _{10}=a+9 d$

Given that, $a_4+a_8=24$

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

$2 a+10 d=24$

$a+5 d=12(1)$

$a_6+a _{10}=44$

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

$2 a+14 d=44$

$a+7 d=22(2)$

On subtracting equation (1) from (2), we obtain

$2 d=22-12$

$2 d=10$

$d=5$

From equation (1), we obtain

$a+5 d=12$

$a+5(5)=12$

$a+25=12$

$a=-13$ $a_2=a+d=-13+5=-8$

$a_3=a_2+d=-8+5=-3$

Therefore, the first three terms of this A.P. are $-13,-8$, and -3 .

Solution

Given that,

$a=5$

$d=1.75$

$a_n=20.75$

$n=?$

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

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

$15.75=(n-1) 1.75$

$(n-1)=\dfrac{15.75}{1.75}=\dfrac{1575}{175}$

$=\dfrac{63}{7}=9$

$n-1=9$

$n=10$

Hence, $n$ is 10 .