MATH

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AN AP has 21 term the sum of 10 ,11,12 term is 129 and sum of last 3 term is 237.find AP

  • MATH -

    T10+T11+T12 = (a+9d)+(a+10d)+(a+11d), so
    3a+30d = 129
    Similarly, for the last three terms,
    3a+57d = 237

    a=3
    d=4

    and the sequence is

    3,7,11,...,83

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