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if the sixth term of an arithmetic progression is 37 and sum of the first six term is 147

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  1. a1 is the initial term of an arithmetic progression.

    The common difference of successive members is d.

    The nth term of an arithmetic progression.

    an = a1 + ( n - 1 ) d

    In this case:

    n = 6

    a6 = a1 + ( 6 - 1 ) d = 37

    a1 + 5 d = 37

    The sum of the n terms of an arithmetic is:

    Sn = n * ( a1 + an ) / 2

    S6 = 6 * ( a1 + a6 ) / 2 = 147

    3 * ( a1 + a6 ) = 147

    3 * ( a1 + a1 + 5 d ) = 147

    3 * ( 2 a1 + 5 d ) = 147

    6 a1 + 15 d = 147

    Now you must solve system:

    a1 + 5 d = 37

    6 a1 + 15 d = 147

    Try that.

    The solutions are:

    a1 = 12 , d = 5

    Proof:

    a1 = 12

    a2 = 12 + 5 = 17

    a3 = 17 + 5 = 22

    a4 = 22 + 5 = 27

    a5 = 27 + 5 = 32

    a6 = 32 + 5 = 37

    S6 = a1 + a2 + a3 + a4 + a5 + a6 =

    12 + 17 + 22 + 27 + 32 + 37 = 147

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  2. the sixth term of an arithmetic progression is 37
    ---> a+5d = 37
    5d = 37-a

    sum of the first six term is 147
    ---> (6/2)(2a + 5d) = 147
    3(2a + 37-a) = 147
    a +37 = 49
    a = 12
    d = 5

    If you had a question, take it from there

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  3. SLIMPLIFY 2 TAN 60+COS30 DIVIDED BY SIN 60

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  4. What may be the appropriate famular for this question?

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