1975 AHSME Problems/Problem 16

Problem

If the first term of an infinite geometric series is a positive integer, the common ratio is the reciprocal of a positive integer, and the sum of the series is $3$ , then the sum of the first two terms of the series is

$\textbf{(A)}\ \frac{1}{3} \qquad \textbf{(B)}\ \frac{2}{3} \qquad \textbf{(C)}\ \frac{8}{3} \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ \frac{9}{2} \qquad$

Solution

Let's establish some ground rules...

$a =$ The first term in the \href{https://artofproblemsolving.com/wiki/index.php/Geometric_sequence}{geometric sequence}. $r =$ The ratio relating the terms of the \href{https://artofproblemsolving.com/wiki/index.php/Geometric_sequence}{geometric sequence}. $n =$ The nth value of the \href{https://artofproblemsolving.com/wiki/index.php/Geometric_sequence}{geometric sequence}, starting at 1 and increasing as consecutive integer values.

Using these terms, the sum can be written as: $sum = a/(1-r) = 3$

Let $x =$ The positive integer that is in the reciprocal of the geometric ratio.

This gives: $3 = a/(1-(1/x))$ $3 = ax/(x-1)$

Now through trial and error we notice that when x = 3 this gives $3 = a(3/2)$ , where $a = 2$ .

Therefore $r = 1/x = 1/3$ . Now we define the sum as $2(1/3)^(n-1)$ .

Now we simply add the $n = 1 and n = 2$ terms.

$sum = 2(1/3)^((1)-1) + 2(1/3)^((2)-1) = 2 + 2/3 = 6/3 + 2/3 = 8/3$

This gives $\boxed{\textbf{(B) } 8/3}$ .

~PhysicsDolphin

1975 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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1975 AHSME Problems/Problem 16

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