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.
My answer is 12276, is this correct?
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,... :-(
the sum of the first 10 terms of the geometric sequence.
Math
You're welcome! If you need further assistance, feel free to ask.
Steve is right. I messed up doubling one term. Sorry about that
12 + 24 + 48 + 96 +192 +384 +768 + 1536 +3072 + 6144 = 12,276
Congratulations!
I used a brute force method. Steve's is better
2+5+8+11+14+17+20......=?
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