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In an AP(viii) Given an = 4, d = 2, Sn = − 14, find n and a. Q.3(8)

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What is the solution of class10 ex.5.3 Q.no.3 (8) please give me easy solution In an AP(viii) Given an = 4, d = 2, Sn = − 14, find n and a

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  1. Given that, nth term, an = 4, common difference, d = 2, sum of n terms, Sn = −14.

    As we know, from the formula of the nth term in an AP,

    an = a+(n −1)d,

    Therefore, putting the given values, we get,

    4 = a+(−1)2

    4 = a+2n−2

    a+2n = 6

    = 6 − 2n …………………………………………. (i)

    As we know, the sum of n terms is;

    Sn = n/2 (a+an)

    -14 = n/2 (a+4)

    −28 = (a+4)

    −28 = (6 −2n +4) {From equation (i)}

    −28 = (− 2n +10)

    −28 = − 2n2+10n

    2n2 −10n − 28 = 0

    n2 −5−14 = 0

    n2 −7n+2n −14 = 0

    (n−7)+2(n −7) = 0

    (n −7)(n +2) = 0

    Either n − 7 = 0 or n + 2 = 0

    n = 7 or n = −2

    However, n can neither be negative nor fractional.

    Therefore, n = 7

    From equation (i), we get

    a = 6−2n

    a = 6−2(7)

    = 6−14

    = −8

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