Maths

The 5th term of an arithmetic progression is 3times of 2nd term and 12th term exceeds 2times of 6th term by 1. Find the 16th term

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asked by Anonymous

  1. a5 = 3 a2

    a12 = 2 a6 + 1


    For AP:

    an = a1 + ( n - 1 ) d

    a1 = initial term of an arithmetic progression

    d = common difference of successive members


    a2 = a1 + ( 2 - 1 ) d

    a2 = a1 + 1 * d

    a2 = a1 + d


    a5 = a1 + ( 5 - 1 ) d

    a5 = a1 + 4 d


    a6 = a1 + ( 6 - 1 ) d

    a6 = a1 + 5 d


    a12 = a1 + ( 12 - 1 ) d

    a12 = a1 + 11 d


    Now:

    a5 = 3 a2

    a5 = 3 ( a1 + d )

    a5 = 3 a1 + 3 d


    a5 = a5

    a1 + 4 d = 3 a1 + 3 d Subtract a1 to both sides

    a1 + 4 d - a1 = 3 a1 + 3 d - a1

    4 d = 2 a1 + 3 d Subtract 3 d to both sides

    4 d - 3 d= 2 a1 + 3 d - 3 d

    d = 2 a1


    a12 = 2 a6 + 1

    a1 + 11 d = 2 ( a1 + 5 d ) + 1

    a1 + 11 d = 2 a1 + 10 d + 1 Subtract a1 to both sides

    a1 + 11 d - a1 = 2 a1 + 10 d + 1 - a1

    11 d = a1 + 10 d + 1 Subtract 10 d to both sides

    11 d - 10 d = a1 + 10 d + 1 - 10 d

    d = a1 + 1


    d = d

    2 a1 = a1 + 1 Subtract 1 to both sides

    2 a1 - 1 = a1 + 1 - 1

    2a1 - 1 = a1 Subtract a1 to both sides

    2a1 - 1 - a1 = a1 - a1

    a1 - 1 = 0 Add 1 to both sides

    a1 - 1 + 1 = 0 + 1

    a1 = 1


    d = 2 a1

    d = 2 * 1

    d = 2


    an = a1 + ( n - 1 ) d

    a16 = a1 + ( 16 - 1 ) d

    a16 = a1 + 15 d

    a16 = 1 + 15 * 2

    a16 = 1 + 30

    a16 = 31

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    posted by Bosnian

  2. a+4d = 3(a+d)
    a+11d = 2(a+5d)+1

    Solve for a and d, then

    T16 = a+15d

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    posted by Steve

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