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

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==Solution 1==
 
==Solution 1==
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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>
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Let <math>a</math> be the first term of the sequence and <math>r</math> be the common ratio.  Then <math>a+ar = a(1+r) = 7</math> and <math>a+ar+ar^2+ar^3+ar^4+ar^5 = a(1+r)(1+r^2+r^4) = 7(1+r^2+r^4) = 91</math>.  So <math>1+r^2+r^4=13</math>, and <math>r^4+r^2-12=(r^2-3)(r^2+4)=0</math>.  Since <math>r</math> is real, <math>r^2 = 3</math>, so the sum of the first <math>4</math> terms is <math>a+ar+ar^2+ar^3 = a(1+r)(1+r^2) = 7\cdot 4 = \boxed{(\textbf{A})\ 28}</math>.
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 +
==Solution 2==
 +
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 {(\textbf{A})\ 28}</math>
  
 
~Arcticturn
 
~Arcticturn

Latest revision as of 18:17, 18 August 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

Let $a$ be the first term of the sequence and $r$ be the common ratio. Then $a+ar = a(1+r) = 7$ and $a+ar+ar^2+ar^3+ar^4+ar^5 = a(1+r)(1+r^2+r^4) = 7(1+r^2+r^4) = 91$. So $1+r^2+r^4=13$, and $r^4+r^2-12=(r^2-3)(r^2+4)=0$. Since $r$ is real, $r^2 = 3$, so the sum of the first $4$ terms is $a+ar+ar^2+ar^3 = a(1+r)(1+r^2) = 7\cdot 4 = \boxed{(\textbf{A})\ 28}$.

Solution 2

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 $4x = 4 \cdot 7 = \boxed {(\textbf{A})\ 28}$

~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

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