Question

Find the sum of two middle most terms of the AP $-\frac{4}{3},-1,\frac{-2}{3},...,4\frac{1}{3}$. [CBSE 2013C]

Answer 1

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

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,

$$\begin{gathered} a_n=4\frac{1}{3} \\ \Rightarrow -\frac{4}{3}+\left(n-1\right)\times \left(\frac{1}{3}\right)=\frac{13}{3}\left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow \frac{1}{3}\left(n-1\right)=\frac{13}{3}+\frac{4}{3}=\frac{17}{3} \\ \Rightarrow n-1=17 \end{gathered}$$
$\Rightarrow n=17+1=18$

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. 9th term and 10th term.

∴ Sum of the middle most terms of the given AP

= 9th term + 10th term

$$\begin{gathered} =\left[-\frac{4}{3}+\left(9-1\right)\times \frac{1}{3}\right]+\left[-\frac{4}{3}+\left(10-1\right)\times \frac{1}{3}\right] \\ =-\frac{4}{3}+\frac{8}{3}-\frac{4}{3}+3 \\ =3 \end{gathered}$$

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