# 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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posted by Jm
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.

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posted by drwls
3. the sum of the first 10 terms of the geometric sequence.

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

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

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posted by drwls
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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posted by drwls
8. 2+5+8+11+14+17+20......=?

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posted by Clement
9. Math

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posted by Richard

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