# 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) $\square, 38, \square, \square, \square, -22$

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To do:

We have to find the missing term in the box in each case.

Solution:

(i) Given AP is $2, \square, 26$

$a=2, a_{3}=26$

We know that,

$a_{n}=a+(n-1) d$

$26=a+2 d$

$26 =2+2 d$

$2 d=26-2$

$2d=24$

$d=\frac{24}{2}$

$d=12$

$a_{2}=a+(2-1) d$

$=a+d$

$=2+12$

$=14$

The missing term in the box is 14.

(ii) Given AP is $\square, 13, \square, 3$

$a=13$

We know that,

$a_{2}=a+(2-1) d$

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

$a_{4}=3$

$a_{4}=a+(4-1) d$

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

Subtracting (i) from (ii), we get,

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

$2d=-10$

$d=\frac{-10}{2}$

$d=-5$

This implies,

$a+d =13$

$a+(-5)=13$

$a=13+5$

$a=18$

$a_{3}=a+2 d$

$=18+2(-5)$

$=18-10$

$=8$

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

$a=5$

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

We know that,

$a_{4}=a+(4-1) d$

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

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

$3d=\frac{19-10}{2}$

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

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

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

This implies,

$a_{2}=a+(2-1) d$

$=a+d$

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

$=\frac{13}{2}$

$a_{3}=a_{2}+d$

$=\frac{13}{2}+\frac{3}{2}$

$=\frac{16}{2}$

$=8$

(iv) Given AP is $-4, \square, \square, \square, \square, 6$

$a=-4, a_{6}=6$

We know that,

$a_{n}=a+(n-1) d$

$a_{2}=a+d$

$=-4+2$

$=-2$

$a_{3}=a_{2}+d$

$=-2+2$

$=0$

$a_{4}=a_{3}+d$

$=0+2$

$=2$

$a_{5}=a_{4}+d$

$=2+2$

$=4$

(v) Given AP is $\square, 38, \square, \square, \square, -22$

$a_{2}=38, a_{6}=-22$

We know that,

$a_{n}=a+(n-1) d$

$a_{2}=a+d$

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

$a_{6}=a+5 d$

$=-22$......(ii)

Subtracting (i) from (ii), we get,

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

$4d=-60$

$d=\frac{-60}{4}$

$d=-15$

This implies,

$a+d=38$

$a=38-(-15)$

$a=38+15$

$a=53$

$a_3=a_2+d$

$=38+(-15)$

$=38-15$

$=23$

$a_4=a_3+d$

$=23+(-15)$

$=23-15$

$=8$

$a_5=a_4+d$

$=8+(-15)$

$=8-15$

$=-7$

Updated on 10-Oct-2022 13:20:19