Question

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

Answer 1

(i) For this A.P.,

$a=2$

$a_3=26$

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

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

$26=2+2d$

$24=2d$

$d=12$

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

$=14$

Therefore, $14$ is the missing term.

(ii) For this A.P.,

$a_2=13$ and

$a_4=3$

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

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

$13=a+d$ ... (i)

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

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

On subtracting (i) from (ii), we get,

$-10=2d$

$d=-5$

From equation (i), we get,

$13=a+(-5)$

$a=18$

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

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

Therefore, the missing terms are $18$ and $8$ respectively.

(iii) For this A.P.,

$a_1=5$ and

$a_4=9\frac{1}{2}$

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

$a_4=5+(4-1)d$

$9\frac{1}{2}=5+3d$

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

$a_2=a+d$

$a_2=5+\frac{3}{2}$

$a_2=\frac{13}{2}$

$a_3=a_2+\frac{3}{2}$

$a_3=8$

Therefore, the missing terms are $6\frac{1}{2}$ and $8$ respectively.

(iv) For this A.P.,

$a=-4$ and

$a_6=6$

We know that,

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

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

$6=-4+5d$

$10=5d$

$d=2$

$a_2=a+d=-4+2=-2$

$a_3=a+2d=-4+2\left(2\right)=0$

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

$a_5=a+4d=-4+4\left(2\right)=4$

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

(v) For this A.P.,

$a_2=38$

$a_6=-22$

We know that

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

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

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

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

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

On subtracting equation (i) from (ii), we get

$-22-38=4d$

$-60=4d$

$d=-15$

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

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

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

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

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