First term of an infinite geometric sequence

[thornluke]

1. The problem statement, all variables and given/known data
The sum of an infinite geometric sequence is 131/2, and the sum of the first three terms is 13. Find the first term.


2. Relevant equations
S∞ = a/(1-r)
Sn = a-arn/(1-r)


3. The attempt at a solution
a/(1-r) = 131/2

a-ar3/(1-r) = 13

2a = 27-27r ........................ 1
a-ar3 = 13-13r..... 2

I'm stuck.

[pc2-brazil]

2a=27-27r (1)
a-ar^3=13-13r (2)

Put 27 in evidence in (1) and put 13 in evidence in (2).

2a=27(1-r) (1)
a-ar^3=13(1-r) (2)

Now divide (2) by (1).

[HallsofIvy]

Oh!! It is the "a" that cancels allowing you to first solve for r. I was too focused on finding a. Very good pc2-brazil!

[thornluke]

2a=27-27r (1)
a-ar^3=13-13r (2)

Put 27 in evidence in (1) and put 13 in evidence in (2).

2a=27(1-r) (1)
a-ar^3=13(1-r) (2)

Now divide (2) by (1).

2a=27-27r (1)
a-ar^3=13-13r (2)

Put 27 in evidence in (1) and put 13 in evidence in (2).

2a=27(1-r) (1)
a-ar^3=13(1-r) (2)

Now divide (2) by (1).

I found that a =9.

[pc2-brazil]

I found that a =9.
That is correct. I suppose you also found that r = 1/3.
Then, the sum of the first three terms is 13:
9 + 3 + 1 = 13
The sum of the infinite geometric sequence is 13.5:
9 + 3 + 1 + 1/3 + ... = \frac{9}{1-\frac{1}{3}}=13.5