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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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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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  1. 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(n1)

     

    ⇒ 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.

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