Difference between revisions of "1981 AHSME Problems/Problem 14"
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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> | 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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+ | ~Arcticturn |
Revision as of 18:57, 20 October 2021
Problem
In a geometric sequence of real numbers, the sum of the first terms is , and the sum of the first terms is . The sum of the first terms is
Solution 1
Denote the sum of the first terms as . Since we know that the sum of the first terms is which is , we have + + = . We can quickly see that = , and therefore, the the sum of the first terms is
~Arcticturn