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 geometric sequence.


$r =$ The ratio relating the terms of the geometric sequence.


$n =$ The nth value of the 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 careful inspection we notice that when x = 3 the equation becomes

$3 =  a(3/2)$, 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 $\boxed{\textbf{(C) } 8/3}$.


~PhysicsDolphin

For more information on geometric sequences: https://artofproblemsolving.com/wiki/index.php/Geometric_sequence

See Also

1975 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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