In the following APs, find the missing terms in the boxes:
$\square, 38, \square, \square, \square, -22$

Given:

Given AP is $\square, 38, \square, \square, \square, -22$

To do:

We have to find the missing terms in the box.

Solution:

 $a_{2}=38, a_{6}=-22$

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,

$a+d=38$

$a=38-(-15)$

$a=38+15$

$a=53$

$a_3=a_2+d$

$=38+(-15)$

$=38-15$

$=23$

$a_4=a_3+d$

$=23+(-15)$

$=23-15$

$=8$

$a_5=a_4+d$

$=8+(-15)$

$=8-15$

$=-7$

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Updated on: 10-Oct-2022

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