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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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