Question

Find the sum of the two middle most terms of the AP : $-\frac{4}{3},-1,\frac{-2}{3},-\frac{1}{3},....,4\frac{1}{3}$.

Answer 1

Here, the first term $a=\frac{-4}{3}$ and common difference $d=\frac{1}{3}$
Let there are n terms of AP.

Therefore, $a_n=a+(n-1)d=4\frac{1}{3}$
$\Rightarrow \frac{-4}{3}+(n-1)\times \frac{1}{3}=\frac{13}{3}$
$\Rightarrow -4+(n-1)=13$
$\Rightarrow n=18$
Since $n=18$, therefore two most middle terms are 9th and 10th.
Therefore,
$a_9+a_{10}=(a+8d)+(a+9d)=2a+17d$
$\Rightarrow 2\times \frac{-4}{3}+17\times \frac{1}{3}=\frac{9}{3}=3$