Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 1)
(→‎See also)
 
(3 intermediate revisions by 2 users not shown)
Line 5: Line 5:
  
 
==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</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
 +
 
 +
==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 $2$ 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

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. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1981_AHSME_Problems/Problem_14&oldid=252663"