1981 AHSME Problems/Problem 14

Problem

In a geometric sequence of real numbers, the sum of the first $2$ terms is $7$ , and the sum of the first $6$ terms is $91$ . The sum of the first $4$ terms is

$\textbf{(A)}\ 28\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 84$

Solution 1

Let $a$ be the first term of the sequence and $r$ be the common ratio. Then $a+ar = a(1+r) = 7$ and $a+ar+ar^2+ar^3+ar^4+ar^5 = a(1+r)(1+r^2+r^4) = 7(1+r^2+r^4) = 91$ . So $1+r^2+r^4=13$ , and $r^4+r^2-12=(r^2-3)(r^2+4)=0$ . Since $r$ is real, $r^2 = 3$ , so the sum of the first $4$ terms is $a+ar+ar^2+ar^3 = a(1+r)(1+r^2) = 7\cdot 4 = \boxed{(\textbf{A})\ 28}$ .

Solution 2

Denote the sum of the first $2$ terms as $x$ . Since we know that the sum of the first $6$ terms is $91$ which is $7 \cdot 13$ , we have $x$ + $xy$ + $xy^2$ = $13x$ because it is a geometric series. We can quickly see that $y$ = $3$ , and therefore, the sum of the first $4$ terms is $4x = 4 \cdot 7 = \boxed {(\textbf{A})\ 28}$

~Arcticturn

1981 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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1981 AHSME Problems/Problem 14

Contents

Problem

Solution 1

Solution 2

See also