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

Here the given AP is

$-\frac{4}{3},-1,-\frac{2}{3},....,4\frac{1}{3}$
$\\ In this AP: {{a}_{1}} = a=-\frac{4}{3} \\ d={{a}_{2}}-{{a}_{1}}=-1-\left( -\frac{4}{3} \right) \Rightarrow =-1+\frac{4}{3}=\frac{-3+4}{3}=\frac{1}{3} \\$

Let there are n terms in AP

The nth term of AP is

$\\4\frac{1}{3} .\\ {{a}_{n}}=4\frac{1}{3} \\ a+(n-1).d=\frac{13}{3} (Using a\textsubscript{n} = a + (n-1)d)\\$

$\\-\frac{4}{3}+(n-1)\times \left( \frac{1}{3} \right)=\frac{13}{3} \\ \left( n-1 \right)\times \left( \frac{1}{3} \right)=\frac{13+4}{3} \\ n - 1 = 17\\ n = 18\\$

Total terms in this AP is 18 and the middlemost terms and 9thand 10th.

$\\a\textsubscript{9} + a\textsubscript{10} = a + (a - 1)d + a + (10 - 1)d (using a\textsubscript{n} = a + (n-1)d)\\ = 2a + 17d\\$

$\\ =2\left( -\frac{4}{3} \right)+17\left( \frac{1}{3} \right)\\=\frac{-8}{3}+\frac{17}{3} \\ {{a}_{9}}+{{a}_{10}}=\frac{-8+17}{3}=\frac{9}{3}=3 \\ a\textsubscript{9} + a\textsubscript{10} = 3\\$