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In the following APs, find the missing terms in the boxes:
$\square, 13, \square, 3$
Given:
Given AP is $\square, 13, \square, 3$
To do:
We have to find the missing term in the box.
Solution:
$a=13$
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,
$a+d =13$
$a+(-5)=13$
$a=13+5$
$a=18$
$a_{3}=a+2 d$
$=18+2(-5)$
$=18-10$
$=8$
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