math

The first and second terms of an exponential sequence (G.P) are respectively the first and third terms of a linear sequence (A.P). The fourth term of the linear sequence is 10 and sum of its first five terms is 60. find (a) the first five terms of the linear sequence and the sum of the first n terms. (b) the sum Sn of the first n terms of the exponential sequence. (c) the limit of Sn for large value of n

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  1. let the GP be
    a , ar, ar^2 , ar^3, ...

    the arithmetic (linear) sequence is
    a , a+d , a+2d , a + 3d , a+4d ..

    but 2nd term of GP= 3rd term of As
    ar = a + 2d *

    also for the AP
    a+3d = 10 **
    and 5a + 10d = 60
    a + 2d = 12 ***

    subtract (**) - (***)
    d = -2 <------------
    then in **
    a - 6 = 10
    a = 16 <------------

    now back in *
    16r = 16 -4
    r = 12/16 = 3/4 <-----------

    the GP is
    16 , 12 , 9 , 27/4 , ...

    a)
    the AP is
    16, 14 , 12 , 10 , 8 ,6 ...

    sum(n) = (n/2)(2a + (n-1)d)
    = (n/2)(32 - 2(n-1))
    = (n/2)(34 - 2n)
    = n(17 - n)

    b) for the GP
    sum(n) = a(1-r^n)/(1-r)
    = 16( 1 - (3/4)^n)/(1-3/4)
    = 64(1 - (3/4)^n)

    sum(all terms) = a/(1-r)
    = 16/(1-3/4)
    = 64

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  2. Sorry, what is the answer to c?

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  3. Reiny u're not answering. Please I need the answer now

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  4. The third term of a g. P. Is63..and the fifth term is 567. Find the sixth term of the progression. Reiny!!!!!

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  5. T3=ar^2=63 ...i
    T5=ar^4=567 ...ii
    diving ii by i
    ar^4/ar^2=567/63
    r^2=9
    r=3

    ar^2=63
    a=63/9
    a=7

    T6=ar^5
    =7*243
    =1701

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  6. -2,6,-28 . HOW THE COMMON RATIO IS -4

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  7. we really appreciate thanks alot,but my comfussion is about question C.

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