Find the sum of two middle most terms of the AP

Solution:

The given $\mathrm{AP}$ is $-\frac{4}{3},-1, \frac{-2}{3}, \ldots, 4 \frac{1}{3}$.

First term, $a=-\frac{4}{3}$

Common difference, $d=-1-\left(-\frac{4}{3}\right)=-1+\frac{4}{3}=\frac{1}{3}$

Suppose there are n terms in the given AP. Then,

$a_{n}=4 \frac{1}{3}$

$\Rightarrow-\frac{4}{3}+(n-1) \times\left(\frac{1}{3}\right)=\frac{13}{3} \quad\left[a_{n}=a+(n-1) d\right]$

$\Rightarrow \frac{1}{3}(n-1)=\frac{13}{3}+\frac{4}{3}=\frac{17}{3}$

$\Rightarrow n-1=17$

$\Rightarrow n=17+1=18$

Thus, the given AP contains 18 terms. So, there are two middle terms in the given AP.

The middle terms of the given AP are $\left(\frac{18}{2}\right)$ th term and $\left(\frac{18}{2}+1\right)$ th term i.e. 9 th term and 10 th term.

∴ Sum of the middle most terms of the given AP

= 9th term + 10th term

$=\left[-\frac{4}{3}+(9-1) \times \frac{1}{3}\right]+\left[-\frac{4}{3}+(10-1) \times \frac{1}{3}\right]$

$=-\frac{4}{3}+\frac{8}{3}-\frac{4}{3}+3$

$=3$

Hence, the sum of the middle most terms of the given AP is 3.