Difference between revisions of "1971 AHSME Problems/Problem 33"

Latest revision as of 19:47, 1 February 2025

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

Solution 3 (Answer Choices)

We can use dimensional analysis to cut down our answer choices. Suppose that each of the terms in the geometric progression is in units of $\text{meters}$ . Then, $S$ should have units of $\text{meters}$ , $S^{\prime}$ units of $\tfrac1{\text{meters}}$ and $P$ units of $\text{meters}^n$ . Therefore, $SS^{\prime}$ is unitless, so we can eliminate options (A) and (C). $S/S^{\prime}$ has units $\text{meters}^2$ , so, to equal $P$ (which has units $\text{meters}^n$ ), the exponent needs to be $\tfrac n2$ . The only remaining answer choice which satsifies this constraint is $\boxed{\textbf{(B) }(S/S^{\prime})^{\frac12n}}$ .

Video Correction by Dr. Xue's Math School

This problem is WRONG. A counterexample is given in the video:

https://youtu.be/0HSKE4ZC_8Y

@@ Line 26: / Line 26: @@
 == Solution 3 (Answer Choices) ==
 We can use [[dimensional analysis]] to cut down our answer choices. Suppose that each of the terms in the geometric progression is in units of <math>\text{meters}</math>. Then, <math>S</math> should have units of <math>\text{meters}</math>, <math>S^{\prime}</math> units of <math>\tfrac1{\text{meters}}</math> and <math>P</math> units of <math>\text{meters}^n</math>. Therefore, <math>SS^{\prime}</math> is unitless, so we can eliminate options (A) and (C). <math>S/S^{\prime}</math> has units <math>\text{meters}^2</math>, so, to equal <math>P</math> (which has units <math>\text{meters}^n</math>), the exponent needs to be <math>\tfrac n2</math>. The only remaining answer choice which satsifies this constraint is <math>\boxed{\textbf{(B) }(S/S^{\prime})^{\frac12n}}</math>.
+== Video Correction by Dr. Xue's Math School ==
+This problem is WRONG. A counterexample is given in the video:
+https://youtu.be/0HSKE4ZC_8Y
 == 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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