Question

Find the middle most terms of the $A.P\frac{-5}{3},\frac{-4}{3},-1,\frac{-2}{3}........4\frac{2}{3}$

Answer 1

Given,

$a=-\frac{5}{3}$

$d=-\frac{4}{3}+\frac{5}{3}=\frac{1}{3}$

$a_n=4\frac{2}{3}=\frac{14}{3}$

Formula,

$a_n=a+(n-1)d$

$\frac{14}{3}=-\frac{5}{3}+(n-1)\frac{1}{3}$

$14=-5+(n-1)$

$\therefore n=20$

Middle terms will be $10th,11th$ terms.

$a_{10}=-\frac{5}{3}+(10-1)\frac{1}{3}$

$=-\frac{5}{3}+3=\frac{4}{3}$

$a_{11}=-\frac{5}{3}+(11-1)\frac{1}{3}$

$=-\frac{5}{3}+\frac{10}{3}=\frac{5}{3}$