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the sum of first 8 terms of an arithmetic progression is 156.the ratio of its 12th term to its 68th term is 1:5.calculate the first term and the fifteenth term.
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  1. sum of 8 terms is 156
    (8/2)(2a + 7d) = 156
    4(2a + 7d) = 156
    2a + 7d = 39

    (a+11d) : (a + 67d) = 1 : 5

    (a+11d)/(a+67d) = 1/5
    5a + 55d = a + 67d
    4a = 12d
    a = 3d

    sub that into 2a+7d=39
    6d + 7d = 39
    13d=39
    d=3 , then in a = 3d ---> a = 9

    first term is 9
    term 15 = a+14d = 9+14(3) = 51

    check:
    sum(8) = (8/2)(18 + 7(3)) = 4(39) = 156 , checks !
    term12 = a+11d = 42
    term 68 = a+67d = 210
    and 42/210 = 1/5 , YEahh!

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  2. Better explain

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  3. I haven't get your response yet please help me

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  4. How many terms has the A.P Whose first term is 15 and the last is 57 given that the common difference is 3

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  5. Maths

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