Math

The third term of an AP is 8 and ninth term AP exceeds three times the third term by 2. Find the sum of its nineteenth terms

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  1. a+2d = 8
    a+8d = 3*8+2

    so, a=2 and d=3

    Now you want

    S19 = 19/2 (2*2 + 18*3)

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  2. Not satisfied.

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  3. T(3)=8 - - - - - - - (1)
    & T(9)=3*T(3)+2 - - - - - - - - - (2)
    Find S(19)

    T(n) = a+(n-1) d
    Hence equation (1) becomes,
    a+(3-1)d=8
    a+2d=8
    a=8-2d - - - - - - - - - - (3)

    Also equation (2) becomes,
    a+(9-1) d = 3×[a+(3-1) d] +2
    a+8d=[3*(a+2d)]+2
    a+8d=(3a+6d)+2
    a+8d=3a+6d+2
    Rearranging,
    8d-6d=3a-a+2
    2d=2a+2
    Dividing by 2 on both sides,
    d=a+1 - - - - - - - - - - - - - - - - - - - (4)
    Substituting (3) in (4) we get,
    d=(8-2d)+1
    d=8-2d+1
    d=9-2d
    3d=9
    d=9/3
    d=3

    Substituting value of d in (3)
    a=8-2(3)
    a=8-6
    a=2

    Hence a=2 & d=3

    We know that, S(n) = n/2 × [2a+(n-1) d]
    S(19) = 19/2 × [2*2+(19-1)*3]
    = 19/2 * [4+18*3]
    = 19/2 * [4+54]
    =19/2 * 58
    =19*29
    =551

    Sum of first 19 terms of grid AP is 551

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