# Thread: Geometric Progressions

1. ## Geometric Progressions

Hello everyone.

I'm working on the following problem:

A geometric progression containing 20 terms has 48 as its fifth term. The ratio of the sum of its first 6 terms to that of its first 3 terms is 9:1. Find the sum of its last 5 terms.

Here's what I've done with it so far:

G.P.

n = 20

T_5 = 48 => a * r^4 = 48 . . . . . . (i)

S_6 : S_3 = 9 : 1

S_6 / S_3 = 9 / 1

S_6 = 9 * S_3

[ a * (1 - r^5) ] / (1 - r) = { [ a * (1 - r^3) ] / (1 - r) } * 9

1 - r^5 = 9 (1 - r^3) . . . . . . (ii)

At this point I'm stuck... can't seem to figure out a way to get a and r.

Once I have a and r the other part of the question is trivial.

Any help will be greatly appreciated.

Thanks,

Shahz.

2. A geometric progression containing 20 terms has 48 as its fifth term. The ratio of the sum of its first 6 terms to that of its first 3 terms is 9:1. Find the sum of its last 5 terms.
$\displaystyle \frac{a+ar+ar^2+ar^3+ar^4+ar^5}{a+ar+ar^2} = \frac{9}{1}$

$\displaystyle 1+r+r^2+r^3+r^4+r^5 = 9+9r+9r^2$

$\displaystyle r^3+r^4+r^5 = 8(1+r+r^2)$

$\displaystyle r^3(1+r+r^2) = 8(1+r+r^2)$

$\displaystyle r = 2$

since $\displaystyle ar^4 = 48$

$\displaystyle a = 3$

3. Originally Posted by skeeter
$\displaystyle \frac{a+ar+ar^2+ar^3+ar^4+ar^5}{a+ar+ar^2} = \frac{9}{1}$

$\displaystyle 1+r+r^2+r^3+r^4+r^5 = 9+9r+9r^2$

$\displaystyle r^3+r^4+r^5 = 8(1+r+r^2)$

$\displaystyle r^3(1+r+r^2) = 8(1+r+r^2)$

$\displaystyle r = 2$

since $\displaystyle ar^4 = 48$

$\displaystyle a = 3$
Gr8!! Thanks!!