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Q. If in an A.P., 3rd term is 18 and 7th term is 30, the sum of its 17 terms is :

Sequences and Series

Solution:

$T_{3} = 18 $
$ \Rightarrow a+2d=18 \quad...\left(1\right)$
$ T_{7} = 30$
$ \Rightarrow a+6d = 30 \quad...\left(2\right)$
$ \left(2\right) - \left(1\right)$ gives $4d = 12 $
$ \Rightarrow d=3 $
$S_{17} = \frac{17}{2}\left[2a+\left(17-1\right)d\right]$
$ = \frac{17}{2}\left[2a+16d\right] $
$= 17\left(a+8d\right)= 17\left(a+6d+2d\right) $
$= 17\left(30+6\right)$
$ = 17\times36 $
$ = 612$