# Find the sum of the two middle most terms

Question:

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

Solution:

Here, first term $(a)=-\frac{4}{3}$, common difference $(d)=-1+\frac{4}{3}=\frac{1}{3}$

and the last term $(l)=4 \frac{1}{3}=\frac{13}{3}$

$\because n$th term of an AP, $l=a_{n}=a+(n-1) d$

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

$\Rightarrow \quad 13=-4+(n-1)$

$\Rightarrow \quad n-1=17$

$\Rightarrow \quad n=18$ [even]

So, the two middle most terms are $\left(\frac{n}{2}\right)$ th and $\left(\frac{n}{2}+1\right)$ th. i.e., $\left(\frac{18}{2}\right)$ th and $\left(\frac{18}{2}+1\right)$ th terms

i.e., 9th and 10th terms.

$\therefore \quad a_{9}=a+8 d=-\frac{4}{3}+8\left(\frac{1}{3}\right)=\frac{8-4}{3}=\frac{4}{3}$

and $a_{10}=a+9 d=\frac{-4}{3}+9\left(\frac{1}{3}\right)=\frac{9-4}{3}=\frac{5}{3}$

So, sum of the two middle most terms $=a_{9}+a_{10}=\frac{4}{3}+\frac{5}{3}=\frac{9}{3}=3$