First term a = \frac{-4}{3}3−4​

Common difference = -1 + \frac{4}{3}34​

= \frac{1}{3}31​

nth term = a(n) = \frac{13}{3}313​

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

∴ nth term of an AP = a + (n – 1) * d

⇒ \frac{13}{3}313​ = \frac{-4}{3}3−4​ + (n – 1) * \frac{1}{3}31​

⇒ \frac{13}{3}313​ + \frac{4}{3}34​ = (n – 1) * \frac{1}{3}31​

⇒ \frac{17}{3}317​ = \frac{(n – 1)}{3}3(n−1)​

⇒ 17 = (n – 1)

⇒ n = 18

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

= (\frac{18}{2}218​ ), ({18}{2}182 ) + 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}3−4​ + 8(\frac{1}{3}31​ )] + [\frac{-4}{3}3−4​ + 9(\frac{1}{3}31​ )]

= [\frac{-4}{3}3−4​ + \frac{8}{3}38​ + [\frac{-4}{3}3−4​ + 3]

= \frac{-4}{3}3−4​ + \frac{8}{3}38​ – \frac{4}{3}34​ + 3

=\frac{-8}{3}3−8​ + \frac{8}{3}38​ + 3

= 3.