Difference between revisions of "1981 AHSME Problems/Problem 14"

(Solution 1)
(Solution 1)
 
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==Solution 1==
 
==Solution 1==
Denote the sum of the first <math>2</math> terms as <math>x</math>. Since we know that the sum of the first <math>6</math> terms is <math>91</math> which is <math>7 \cdot 13 = 91</math>, we have <math>x</math> + <math>xy</math> + <math>xy^2</math> = <math>13x</math>. We can quickly see that <math>y</math> = <math>3</math>, and therefore, the the sum of the first <math>4</math> terms is <math>4x = 4 \cdot 7 = \boxed {(A) 28}</math>
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Denote the sum of the first <math>2</math> terms as <math>x</math>. Since we know that the sum of the first <math>6</math> terms is <math>91</math> which is <math>7 \cdot 13</math>, we have <math>x</math> + <math>xy</math> + <math>xy^2</math> = <math>13x</math> because it is a geometric series. We can quickly see that <math>y</math> = <math>3</math>, and therefore, the sum of the first <math>4</math> terms is <math>4x = 4 \cdot 7 = \boxed {(A) 28}</math>
  
 
~Arcticturn
 
~Arcticturn

Latest revision as of 20:00, 20 October 2021

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