1981 AMC 12 第 14 题

Question

一个实数等比数列的前两项之和为

7

，前六项之和为

91

。它的前四项之和为

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

把前六项之和分解成前两项之和与

1+r^2+r^4

的乘积

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

1+r^2+r^4

令

t=r^2

，解方程

1+t+t^2=13

Set

t=r^2

and solve

1+t+t^2=13

若首项为

a

，公比为

r

，则

91=a(1+r)(1+r^2+r^4)\text{。}

由于

a(1+r)=7

，可得

1+r^2+r^4=13

。令

t=r^2\ge0

，则

t^2+t-12=0

，所以

t=3

。前四项之和为

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

。

所以正确答案是 A。

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,

so

t=3.

The first four terms sum to

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

Therefore, the correct answer is A.

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