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Rajan@2021
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How many terms of the AP 63, 60, 57, 54,….must be taken so that their sum is 693? Explain the double answer.

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One of the basic question from arithmetic progression chapter in which we have been asked to find the number of terms required so that the sum must be 693 of the given series.

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

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1 Answer

  1. Consider
    a= First term
    d= Common difference
    n= number of terms

    Here, a=63,d=6063=3 and sum =Sn=693

    Sn=n/2[2a+(n1)d]

    693=n/2[2×63+(n1)(3)]
    693×2=n(1263n+3)
    1386=n(1293n)
    1386=129n3n2
    3n^2129n+1386=0
    n^243n+462=0
    Which is a quadratic equation
    n^221n22n+462=0
    n(n21)22(n21)=0
    (n21)(n22)=0
    Either, n21=0, then n=21
    or n22=0, then n=22

    Number of terms=21 or 22
    T22=a+(n1)d
    =63+(221)(3)
    =63+21×(3)=6363=0
    Which shows that, 22th term of AP is zero.
    Number of terms are 21 or 22. So there will be no effect on the sum.

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