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

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 $4x = 4 \cdot 7 = \boxed {(A) 28}$

~Arcticturn

1981 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
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1981 AHSME Problems/Problem 14

Problem

Solution 1

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