Question

Question 19
Find the sum of the two middle most terms of the AP $-\frac{4}{3},-1,-\frac{2}{3},\cdots ,4\frac{1}{3}$

Answer 1

Here, first term $\left(a\right)=-\frac{4}{3}$, common difference (d) = $-1+\frac{4}{3}=\frac{1}{3}$
And the last term $\left(I\right)=4\frac{1}{3}=\frac{13}{3}$
nth term of an AP, $I=a_n=a+(n-1)d$
$\Rightarrow \frac{13}{3}=-\frac{4}{3}+(n-1)\frac{1}{3}$
$\Rightarrow 13=-4+(n-1)$
$\Rightarrow n-1=17$
$\Rightarrow n=18$
So, the two middle most terms are
$\left(\frac{n}{2}\right)thand(\frac{n}{2}+1)thi.e.,\left(\frac{18}{2}\right)thand(\frac{18}{2}+1)th$ terms i.e., $9^{th}$ and ${10}^{th}$ terms.
$\therefore a_9=a+8d=-\frac{4}{3}+8\left(\frac{1}{3}\right)=\frac{8-4}{3}=\frac{4}{3}$
$And{a}_{10}=a+9d=-\frac{-4}{3}+9\left(\frac{1}{3}\right)=\frac{9-4}{3}=\frac{5}{3}$
So, sum of the two middle most terms $=a_9+a_{10}=\frac{4}{3}+\frac{5}{3}=\frac{9}{3}=3$