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Find the sum of two middle terms of the A.P.:$-\frac{4}{3}, -1, -\frac{2}{3}, -\frac{1}{3}, ......, 4\frac{1}{3}$.
Given:
Given A.P. is $-\frac{4}{3}, -1, -\frac{2}{3}, -\frac{1}{3}, ......, 4\frac{1}{3}$.
To do:
We have to find the sum of the two middle terms of the given A.P.
Solution:
$a_1=a=-\frac{4}{3}, a_2=-1, l=4\frac{1}{3}$
Common difference $d=-1-(-\frac{4}{3})=-1+\frac{4}{3}=\frac{-1(3)+4}{3}=\frac{1}{3}$
Let there be $n$ terms in the given A.P.
This implies,
$l=a_n=-\frac{4}{3}+(n-1)(\frac{1}{3})$
$4\frac{1}{3}=-\frac{4}{3}+(n-1)\frac{1}{3}$
$\frac{3\times4+1}{3}=\frac{-4+(n-1)}{3}$
$12+1=-4+n-1$
$n=13+5$
$n=18$
Here, $n=18$ is even.
Therefore, $(\frac{n}{2})$th and $\frac{n}{2}+1$th terms are the middle terms.
$\frac{n}{2}=\frac{18}{2}=9$
$\frac{n}{2}+1=9+1=10$
Middle terms are $a_{9}$ and $a_{10}$
$a_{9}=-\frac{4}{3}+(9-1)(\frac{1}{3})$
$=\frac{-4+8}{3}$
$=\frac{4}{3}$
$a_{10}=a_9+d=\frac{4}{3}+\frac{1}{3}=\frac{4+1}{3}=\frac{5}{3}$
$a_9+a_{10}=\frac{4}{3}+\frac{5}{3}=\frac{4+5}{3}=\frac{9}{3}=3$
The sum of the middle terms of the given A.P. is $3$.
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