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Rajan@2021
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Solve the equation : 1+4+7+10+…+x=287

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An important question from ML Aggarwal(avichal Publication).  We have been asked to find the value of X.

Solve the equation : 1+4+7+10+…+x=287

Arithmetic Progression, Chapter 9, Question no 6

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1 Answer

  1. Here,givena=1

    d=41=3

    and,sn=287

    Now,

    sn=2n(2a+(n1)d)

    287=2n(2×1+(n1)3)

    287=2n(2+3n3)

    574=n(3n1)

    574=3n2n3n2n574=0

    on solving the quadratic equaton using formula n=2ab±b24ac

    We get n=14 & 3−41[does not exist] so, n=14

    Now,

    sn=2n(a+1)

    287=214(1+x)

    574=14(1+x)

    (1+x)=145741+x=41

    x=411

    x=40

    x=40 is the solution.

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