Question

Find the sum of two middle most terms of the AP -4/3,-1,-2/3,.....,4•1/3​

Answer 1

$\huge\bold\star{\fcolorbox{black}{pink}{Answer}}$

3

$\sf\green{\underline{\sf Given, }}$

First term a = $\frac{-4}{3}$

Common difference = -1 + $\frac{4}{3}$

= $\frac{1}{3}$

nth term = a(n) = $\frac{13}{3}$

$\sf\orange{\underline{\sf Solution :}}$

∴ nth term of an AP = a + (n - 1) * d

⇒ $\frac{13}{3}$ = $\frac{-4}{3}$ + (n - 1) * $\frac{1}{3}$

⇒ $\frac{13}{3}$ + $\frac{4}{3}$ = (n - 1) * $\frac{1}{3}$

⇒ $\frac{17}{3}$ = $\frac{(n - 1)}{3}$

⇒ 17 = (n - 1)

⇒ n = 18

∴ So, the given ap contains 18 terms. The middle terms are ($\frac{n}{2}$), ($\frac{n}{2}$) + 1.

= ($\frac{18}{2}$), (${18}{2}$) + 1

= 9,10

Sum of two middle terms:

= 9th term + 10th term

= [a + (9 - 1) * d] + [a + (10 - 1) * d]

= [a + 8d] + [a + 9d]

= $\frac{-4}{3}$ + 8($\frac{1}{3}$)] + [$\frac{-4}{3}$ + 9($\frac{1}{3}$)]

= [$\frac{-4}{3}$ + $\frac{8}{3}$] + [$\frac{-4}{3}$+ 3]

= $\frac{-4}{3}$ + $\frac{8}{3}$ - $\frac{4}{3}$ + 3

=$\frac{-8}{3}$ + $\frac{8}{3}$ + 3

= 3.

⁂︎ ,Sum of two middle most terms is 3.