1981 AMC 12 Problem 14

Attempt Problem 14 of the 1981 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1981 AMC 12 solutions, or check the answer key.

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14.

In a geometric sequence of real numbers, the sum of the first two terms is

7,

and the sum of the first six terms is

91.

The sum of the first four terms is

28

32

35

49

84

Factor the first-six-term sum into the first-two-term sum times

1+r^2+r^4

Set

t=r^2

and solve

1+t+t^2=13

If the first term is

a

and ratio is

r,

then

91=a(1+r)(1+r^2+r^4).

Since

a(1+r)=7,

we have

1+r^2+r^4=13.

With

t=r^2\ge0,

t^2+t-12=0,

t=3.

The first four terms sum to

a(1+r)(1+r^2)=7(4)=28.

Therefore, the correct answer is A.

Problem 14 in other years