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1971 AHSME Problems/Problem 33

Revision as of 18:12, 8 August 2024 by Thepowerful456 (talk | contribs) (solution that doesn't rely on answer choices, see also box)

Problem

If $P$ is the product of $n$ quantities in Geometric Progression, $S$ their sum, and $S'$ the sum of their reciprocals, then $P$ in terms of $S, S'$, and $n$ is

$\textbf{(A) }(SS')^{\frac{1}{2}n}\qquad \textbf{(B) }(S/S')^{\frac{1}{2}n}\qquad \textbf{(C) }(SS')^{n-2}\qquad \textbf{(D) }(S/S')^n\qquad \textbf{(E) }(S/S')^{\frac{1}{2}(n-1)}$

Solution 1

Let the geometric sequence have first term $a$ and common ratio $R$. Then, the first $n$ terms of the sequence are $a,aR,aR^2,\ldots,aR^{n-1}$. The product of these terms $P$ is $a^nR^{1+2+\ldots+n-1}=a^nR^{\frac{(n-1)n}2}$ by the formula for triangular numbers. Using the sum formula reveals that $S=a\cdot\frac{1-R^n}{1-R}$.

We know that $S^{\prime}=\tfrac1a+\tfrac1{aR}+\ldots+\tfrac1{aR^{n-1}}$ Combining fractions reveals that $S^{\prime}=\frac{aR^{n-1}+aR^{n-2}+\ldots+aR+a}{a^2R^{n-1}}=\frac S{a^2R^{n-1}}$. Note that this denominator looks suspiciously similar to our formula for $P$. In fact, $(a^2+R^{n-1})^{\frac n2}=a^n+R^{\frac{(n-1)n}2}=P$. Because $(S/S^{\prime})^{\frac n2}=(a^2+R^{n-1})^{\frac n2}=P$, our answer is $\boxed{\textbf{(B) }(S/S^{\prime})^{\frac 12n}}$.

Solution 2 (Answer Choices)

We can just look at a very specific case: $1, 2, 4, 8.$ Here, $n=4, P=64, S=15,$ and $S'=\frac{30}{16}=\frac{15}{8}.$

Then, plug in values of $S, S',$ and $n$ into each of the answer choices and see if it matches the product.

Answer choice $\boxed{\textbf{(B) }(S/S')^{\frac{1}{2}n}}$ works: $(\frac{15}{\frac{15}{8}})^2 = 64.$

-edited by coolmath34

See Also

1971 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 32 		Followed by
Problem 34
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 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

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