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Rajan@2021
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In an AP, it is given that S5+S7 =167 and S10 =235, then find the AP, where Sn denotes the sum of its first n terms.

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I came across a difficult question from arithmetic progression chapter and it was already asked in previous year paper of 2015 in which we have to find the sum of the first n terms if in an AP, it is given that S5+S7 =167 and S10 =235, where Sn denotes the sum of its first n terms.

Book – RS Aggarwal, Class 10, chapter 5C, question no 20.

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1 Answer

  1. Let the first term is a and the common difference is d

    By using Sn​=n/2​[2a+(n−1)d] we have,

    S5​=5/2​[2a+(5−1)d]=5/2[2a+4d]

    S7​=7/2​[2a+(7−1)d]=7/2​[2a+6d]

    Given: S7​+S5​=167

    ∴5/2​[2a+4d]+7/2​[2a+6d]=167

    ⇒10a+20d+14a+42d=334

    ⇒24a+62d=334       …(1)

    S10​=10/2​[2a+(10−1)d]=5(2a+9d)

    Given: S10​=235

    So 5(2a+9d)=235

    ⇒2a+9d=47       …(2)

    Multiply equation (2) by 12, we get

    24a+108d=564....(3)

    Subtracting equation (3) from (1), we get

    −46d=−230

     ∴d=5

    Substing the value of d=5 in equation (1) we get

    2a+9(5)=47
    or 2a=2

     ∴a=1

    Then A.P is 1,6,11,16,21,⋯

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