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How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636? Q.4

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What is the solution of question 4 of class 10 arithmetic progressions. Best and easy way to solve the question its very important for class 10. How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?

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  1. Let there be n terms of the AP. 9, 17, 25 …

    For this A.P.,

    First term, a = 9

    Common difference, d = a2a1 = 17−9 = 8

    As, the sum of n terms, is;

    Sn = n/2 [2a+(n -1)d]

    636 = n/2 [2×a+(8-1)×8]

    636 = n/2 [18+(n-1)×8]

    636 = [9 +4n −4]

    636 = (4n +5)

    4n2 +5n −636 = 0

    4n2 +53n −48n −636 = 0

    (4n + 53)−12 (4n + 53) = 0

    (4n +53)(n −12) = 0

    Either 4n+53 = 0 or n−12 = 0

    n = (-53/4) or n = 12

    cannot be negative or fraction, therefore, n = 12 only.

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