The sum of an infinite geometric series with common ratio r such that |r|<1 is 15, and the sum of the squares of the terms of this series is 45. The first term of the series is:

1970 AMC 12 Problem 19

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19.

The sum of an infinite geometric series with common ratio

r

such that

|r|\lt1

15,

and the sum of the squares of the terms of this series is

45.

The first term of the series is:

12

10

5

3

2

If the first term is

a,

write one equation for each infinite sum

Use

\frac{a}{1-r}=15

and

\frac{a^2}{1-r^2}=45

, then cancel a factor of

1-r

Let the first term be

a.

Then

\frac{a}{1-r}=15,\qquad \frac{a^2}{1-r^2}=45.

Since

1-r^2=(1-r)(1+r),

dividing the second equation by the first gives

\frac{a}{1+r}=3.

Thus

a=15(1-r)=3(1+r).

Solving gives

r=\frac{2}{3}

and

a=5.

Therefore, the correct answer is C.