Difference between revisions of "1981 AHSME Problems/Problem 14"
Arcticturn (talk | contribs) (Created page with "==Problem== In a geometric sequence of real numbers, the sum of the first <math>2</math> terms is <math>7</math>, and the sum of the first <math>6</math> terms is <math>91</ma...") |
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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</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> | ||
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| + | ~Arcticturn | ||
Latest revision as of 19:00, 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 + 
+ 
= 
because it is a geometric series. We can quickly see that 
= 
, and therefore, the sum of the first terms is 
~Arcticturn
