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Deepak Bora
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Which term of the A.P. 3, 15, 27, 39… will be 132 more than its 54th term?

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The question is given from ncert Book of class 10th Ch. no. 5 Ex. 5.2 Q. 11. In the following question you have to find the term which will be 132 more than the 54th term of the given A.P. Give the solution.

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  1. Solution:-

    It is given in this A.P. that 3, 15, 27, 39, …

    a = 3

    d = a2 − a1 = 15 − 3 = 12

    We know:-

    an = a+(n−1)

    Therefore,

    a54 = a+(54−1)d

    3+(53)(12)

    3+636 = 639

    a54 = 639

    We have to find the term of this series or A.P which is 132 more than a54  Therefore 771.

    Let the nth term be 771.

    an = a+(n−1)d

    771 = 3+(n −1)12

    768 = (n−1)12

    (n −1) = 64

    n = 65

    Hence, the 65 term was 132 more than 54th term.

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