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</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 | ||
| + | |||
| + | ==See also== | ||
| + | |||
| + | {{AHSME box|year=1981|num-b=13|num-a=15}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 14:02, 28 June 2025
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
See also
| 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 | ||
| All AHSME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
