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In an arithmatic sequence, sum of 9 terms is 279 and sum of 20 terms is 1280.
Find 5th term ?
Find 16th term ?
Write the sequence.

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  1. Sum of a certain number of terms of an arithmetic sequence:

    Sn = n ( a1 + an ) / 2

    In this case:

    S9 = 9 ( a1 + a9 ) / 2 = 279

    S20 = 20 ( a1 + a20 ) / 2 = 1280

    9 ( a1 + a9 ) / 2 = 279 Multiply both sides by 2

    9 ( a1 + a9 ) = 279 * 2

    9 a1 + 9 a9 = 558

    On the other side:

    an = a1 ( n - 1 ) d

    in this case: n = 9, n - 1 = 8

    a9 = a1 + 8 d

    9 a1 + 9 a9 = 558

    9 a1 + 9 ( a1 + 8 d ) = 558

    9 a1 + 9 a1 + 9 * 8 d = 558

    9 a1 + 9 a1 + 72 d = 558

    18 a1 + 72 d = 558

    20 ( a1 + a20 ) / 2 = 1280 Multiply both sides by 2

    20 ( a1 + a20 ) = 1280 * 2

    20 a1 + 20 a20 = 2560

    an = a1 ( n - 1 ) d

    In this case: n = 20, n - 1 = 19

    a20 = 20 a1 + 19 d

    20 a1 + 20 a20 = 2560

    20 a1 + 20 ( a1 + 19 d ) = 2560

    20 a1 + 20 a1 + 20 * 19 d = 2560

    20 a1 + 20 a1 + 380 d = 2560

    40 a1 + 380 d = 2560

    Now you must solve system of 2 equations with 2 unknows:

    18 a1 + 72 d = 558

    40 a1 + 380 d = 2560

    The solutions are :

    a1 = 7

    d = 6

    an = a1 + ( n - 1 ) d

    a5 = 7 + ( 5 - 1 ) * 6

    a5 = 7 + 4 * 6

    a5 = 7 + 24

    a5 = 31

    an = a1 + ( n - 1 ) d

    a16 = 7 + ( 16 - 1 ) * 6

    a16 = 7 + 15 * 6

    a16 = 7 + 90

    a16 = 97

    Your sequence:

    7, 13, 19 , 25, 31, 37, 43, 49, 55, 61, 67, 73, 79, 85, 91, 97, 103, 109, 115, 121

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  2. S9 = 279
    F5 = 279/9 = 31
    S20 = 1280
    S20-S9 = 1280-279 =1001
    =[sum of F11 to F20]
    F15 =1001/11 = 91.
    F15-F5 = 91-31 = 60 = 10 d
    d = 60/10 = 6
    F16 = 91 + 6 =97
    F1 =31-4d =31-24= 7
    so, 7,13,19,25,..........

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