Can someone help me?!

The 1st, 5th and 13th terms of an arithmetic sequence are the first three terms of a geometric sequence with a common ratio 2. If the 21st term of the arithmetic sequence is 72, calculate the sum of the first 10 terms of the geometric sequence.

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  1. My answer is 12276, is this correct?

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  2. The arithmetic sequence could be, and is,
    12,15,18,21,24,27,30,33,36,39,
    42,45,48,51,54,57,60,63,66,69,72
    The A1, A5 and A13 terms are 12, 24 and 48. Note that they have the common ratio 2, as required.

    The geometric sequence is
    12, 24, 36, 72, 144, 288, 576, 1152, 2304, 4608.

    It looks like your answer is too large.

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  3. the sum of the first 10 terms of the geometric sequence.

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  4. I just listed all ten for you. Add them up.

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  5. or, analytically,

    a+4d = 2a
    a+12d = 4a

    a = 4d, so, since a+20d=72
    24d=72, and d=3
    so, a=12

    Tn = 12*2^(n-1)
    S10 = 12*(2^10-1)/(2-1) = 12*1023 = 12276

    Looks like you're right.

    drwls, I think the sequence is

    12,24,48,96,... :-(

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  6. Steve is right. I messed up doubling one term. Sorry about that

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  7. 12 + 24 + 48 + 96 +192 +384 +768 + 1536 +3072 + 6144 = 12,276

    Congratulations!

    I used a brute force method. Steve's is better

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  8. 2+5+8+11+14+17+20......=?

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  9. Math

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  10. a=a
    a+4d=ar
    a+12d=ar^2
    an=a+(n-1)d
    72=a+(21-1)d
    72=a+20d............(i)
    a+12d/a+4d=a+4d/a this is because the ratio is same
    a(a+12d)=(a+4d)(a+4d)
    a^2+12da=a^2+4da+4da+16d^2
    a^2+12da=a^2+8da+16d^2
    4da=16d^2
    4a=16d
    a=16d/4
    a=4d................(ii)

    72=4d+20d
    72=24d
    d=3
    hence a=4*3=12

    GP sum=[a(1-r^n)]/1-r
    [12(1-2^10)]/1-2
    [12*(1-1024)]/-1
    12276 ans

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