Question

Find the sum of two middle terms of the A.P. : $-\frac{4}{3},-1,\frac{-2}{3},\frac{-1}{3},.......,4\frac{1}{3}$.

Answer 1

$$\begin{gathered} a=-\frac{4}{3},d=\frac{1}{3},a_n=\frac{13}{3} \\ \Rightarrow a+\left(n-1\right)d=\frac{13}{3} \\ \Rightarrow \left(-\frac{4}{3}\right)+\left(n-1\right)\left(\frac{1}{3}\right)=\frac{13}{3} \\ \Rightarrow 13=-4+n-1 \\ \Rightarrow n=18 \\ \mathrm{Midlle}\mathrm{terms}\mathrm{are}{\frac{\mathrm{n}}{2}}^{\mathrm{th}}\mathrm{and}\frac{\mathrm{n}}{2}+1^{\mathrm{th}},\mathrm{i}.\mathrm{e}9\mathrm{th}\mathrm{and}10\mathrm{th}\mathrm{terms}. \\ {a}_9=a+8d=-\frac{4}{3}+\frac{8}{3}=\frac{4}{3} \\ {a}_{10}=a+9d=-\frac{4}{3}+\frac{9}{3}=\frac{5}{3} \\ \therefore a_9+a_{10}=\frac{9}{3}=3 \end{gathered}$$