Posted by Sniper on Sunday, February 27, 2011 at 3:38am.
1) The sum of last three terms of GP having n terms is 1024 time the sume of first 3 terms of GP. If 3rd term is 5. Find the last term.
2) The sum of the first eight terms of a GP is five times the sum of the first four terms. Find the common ratio.

Geometric  agrin04, Sunday, February 27, 2011 at 4:36am
(1) ar^(n1) + ar^(n2) + ar^(n3) = 1024 (a + ar + ar^2)
ar^(n3) (1 + r + r^2) = 1024a (1 + r + r^2)
r^(n3) = 1024
It is known that the third term is 5, so: ar^2 = 5
The last term will be:
ar^(n1) = ar^2. r^(n3) = 5 x 1024 = ? (You can finish the rest of it, right?)
(2) S8 = 5 S4
a(r^8  1)/(r1) = 5a(r^4  1)/(r  1)
(r^4  1)(r^4 + 1) = 5(r^4  1)
r^4 + 1 = 5
r^4 = 4
r = ±sqrt(2)
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