Question

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

Answer 1

The AP with the first term is $-\frac{4}{3}$ and the last term is $\frac{13}{3}$
The nth term of the AP is given by $a_n=a+(n-1)d$. Here a is the first term and d is the common difference.
Tn is the last term.
$d=T_{n+1}-T_n$,
where n is the natural number.
So,
$d=-1-(-\frac{4}{3})$
common difference $=\frac{1}{3}$
So,
using this formula -----
$a_n=a+(n-1)d$ we get
⇒ $-\frac{4}{3}+(n-1)\frac{1}{3}=\frac{13}{3}$
⇒ $\frac{-5+n}{3}=\frac{13}{3}$
$3n=39+15$
$3n=54$
$n=18$
So, there are 18 terms in the given AP.
The two middlemost terms are the 9th term and 10th term.
Hence, $T_9=-\frac{4}{3}+(9-1)\frac{1}{3}$
$T_9=\frac{4}{3}$
$T_{10}=-\frac{4}{3}+(10-1)\frac{1}{3}$
$T_{10}=\frac{5}{3}$
So, sum of the two middle most terms of the given AP $=\frac{4}{3}+\frac{5}{3}$
$=\frac{9}{3}=3$