# Geometric progression

1. Nov 7, 2006

### gunblaze

ok, the qn goes like this..

Find the first term and common ration of a geometric progression if the sum to n is given by
6 - 2/(3^(n-1))

I tried solving by making both terms of the eqn having the same denominator by multiplying (3^(n-1)) to the first term 6 and then by taking out the 2, i am able to make the formula to 2(3^n - 1)/ (3^n-1). But what about the denominator, i remember that there must be no n at the bottom. Only r - 1 where in this case, r is to be found..

Anyone out here can give me some guide? Thanks for any help given.

2. Nov 8, 2006

### HallsofIvy

Staff Emeritus
First, you have written that wrong. The denominator is not 3^n- 1, it is 3^(n-1). You should have
$$\frac{6(3^{n-1})- 2}{3^{n-1}}= 2(3^n- 1}{3^{n-1}$$

But it's really not necessary to use a general formula. Since you only need to find two numbers, a and r for the general term arn, you really only need two equations. The first term, with n= 1, is a. The formula you are given says it must be
$$a= 6- \frac{2}{3^{1-1}}$$
The sum of the first two terms, with n= 2 is a+ ar and so
$$a+ ar= 6- \frac{2}{3^{2-1}}$$
Can you solve those two equations for a and r?