an important question from the book -ML aggarwal( avichal publication)class 10, Arithmetic Progression……..

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

Arithmetic Progression Chapter 9, ML Publication Class10

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

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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.