Question

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

Answer 1

$\Rightarrow$$a=\frac{-4}{3}$

$\Rightarrow$$d=-1+\frac{4}{3}=\frac{1}{3}$

$\Rightarrow$$l=a+(n-1)d$

$\Rightarrow$$\frac{13}{3}=\frac{-4}{3}+(n-1)\frac{1}{3}$

$\Rightarrow$$\frac{18}{3}=\frac{n}{3}$

$\Rightarrow$$n=18$

middle term$=\frac{18}{2}=9$ and $10$

$\Rightarrow$$T_9=\frac{-4}{3}+(9-1)\frac{1}{3}=\frac{4}{3}$

$\Rightarrow$$T_{10}=\frac{-4}{3}+(10-1)\frac{1}{3}=\frac{5}{3}$

sum of two middle terms$\frac{5}{3}+\frac{4}{3}=\frac{9}{3}=3$