# In the following APs,

Question.

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

(i) $2, \square, 26$

(ii) $\square, 13, \square, 3$

(iii) $5, \square, \square, 9 \frac{1}{2}$

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

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

Solution:

(i) $\mathrm{a}=2, \mathrm{a}+2 \mathrm{~d}=26 \quad \Rightarrow 2+2 \mathrm{~d}=26$

$\Rightarrow 2 \mathrm{~d}=26-2=24 \quad \Rightarrow \mathrm{d}=12$

Then the missing term

$t_{2}=a+d=2+12=14$

(ii) $a+d=13$ ...(1)

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

Subtracting (1) from (2), we get

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

$\Rightarrow 2 \mathrm{~d}=-10 \Rightarrow \mathrm{d}=-5$

from $(1), a-5=13$

$\Rightarrow a=18$

Therefore, the first missing term is 18

The next missing term

$t_{3}=t_{2}+d=13+(-5)=8$

(iii) $a=5$

$a_{4}=9 \frac{1}{2}=\frac{19}{2} \quad a+3 d=\frac{19}{2}$

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

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

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

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

$a_{2}=a+d=5+\frac{3}{2}=\frac{13}{2}$

$a_{3}=a+2 d=5+2\left(\frac{3}{2}\right)=8$

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

(iv) $a=-4$

$a_{6}=6$

$a+5 d=6$

$6=-4+5 \mathrm{~d}$

$10=5 \mathrm{~d}$

$\mathrm{d}=2$

$a_{2}=a+d=-4+2=-2$

$a_{3}=a+2 d=-4+2(2)=0$

$a_{4}=a+3 d=-4+3(2)=2$

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

(v) $\mathrm{a}_{2}=38$

$a_{6}=-22$

$38=a+d$ ...(1)

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

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

$-22-38=4 d$

-60=4 d

$\mathrm{d}=-15$

$a=a_{2}-d=38-(-15)=53$

$a_{2}=a+2 d=53+2(-15)=23$

$a_{4}=a+3 d=53+3(-15)=8$

$a_{5}=a+4 d=53+4(-15)=-7$

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