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^(n-1) + ar^(n-2) + ar^(n-3) = 1024 (a + ar + ar^2)

ar^(n-3) (1 + r + r^2) = 1024a (1 + r + r^2)

r^(n-3) = 1024

It is known that the third term is 5, so: ar^2 = 5

The last term will be:

ar^(n-1) = ar^2. r^(n-3) = 5 x 1024 = ? (You can finish the rest of it, right?)

(2) S8 = 5 S4

a(r^8 - 1)/(r-1) = 5a(r^4 - 1)/(r - 1)

(r^4 - 1)(r^4 + 1) = 5(r^4 - 1)

r^4 + 1 = 5

r^4 = 4

r = ħsqrt(2)

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