Question

The $4^{th}$ term of an AP is 9 and the $9^{th}$ term of the AP is 34. The sum of the first 10 terms of the AP is
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Answer 1

Given, the $4^{th}$ term is 9 and the
$9^{th}$ term is 34 in an AP
If the first term is taken as $a$, and the common difference is taken as $d$, then
$\Rightarrow a_4=a+(4-1)d=9$
$\Rightarrow a_9=a+(9-1)d=34$
From the above equations, we get
$d=5$ and $a=-6$.
Now, sum of the AP upto 10 terms $=S_{10}$
$\Rightarrow S_{10}=\frac{10}{2}\left(2\right(-6)+(10-1\left)5\right)=5(-12+45)=165$