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Difference between revisions of "1955 AHSME Problems/Problem 45"

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== Solution ==  
 
== Solution ==  
 
Let our geometric sequence be <math>a,ar,ar^2</math> and let our arithmetic sequence be <math>0,d,2d</math>. We know that <cmath>\begin{cases} a+0=1 \\ ar+d=1\\ ar^2+2d=2\end{cases}</cmath>
 
Let our geometric sequence be <math>a,ar,ar^2</math> and let our arithmetic sequence be <math>0,d,2d</math>. We know that <cmath>\begin{cases} a+0=1 \\ ar+d=1\\ ar^2+2d=2\end{cases}</cmath>
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This implies that <math>a=1</math>, hence <math>r+d=1</math> and <math>r^2+2d=2</math>. We can rewrite <math>r+d=1</math> as <math>d = 1-r,</math> plugging this into <math>r^2+2d=2,</math> we get <math>r^2+2-2r = 2</math>, we can simplify this to get <math>r(r-2)=0</math>, so <math>r=0</math> or <math>2</math>. But since <math>r\neq 0</math>, <math>r=2</math> and <math>d=-1</math>.  
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This implies that <math>a=1</math>, hence <math>r+d=1</math> and <math>r^2+2d=2</math>. We can rewrite <math>r+d=1</math> as <math>d = 1-r</math>, plugging this into <math>r^2+2d=2</math>, we get <math>r^2+2-2r = 2</math>, we can simplify this to get <math>r(r-2)=0</math>, so <math>r=0</math> or <math>2</math>. But since <math>r\neq 0</math>, <math>r=2</math> and <math>d=-1</math>.  
 
So our two sequences are <math>\{1-n\}</math> and <math>\{2^{n-1}\}</math>, which means the third sequence will be <cmath>\{2^{n-1}-n+1\}=\{1,1,2,5,12,27,\dots\}</cmath>Calculating the sum of the first 10 terms and adding them up yields 978, hence our answer is <math>\fbox{A}</math>.
 
So our two sequences are <math>\{1-n\}</math> and <math>\{2^{n-1}\}</math>, which means the third sequence will be <cmath>\{2^{n-1}-n+1\}=\{1,1,2,5,12,27,\dots\}</cmath>Calculating the sum of the first 10 terms and adding them up yields 978, hence our answer is <math>\fbox{A}</math>.

Revision as of 16:22, 2 July 2020

Problem 45

Given a geometric sequence with the first term $\neq 0$ and $r \neq 0$ and an arithmetic sequence with the first term $=0$. A third sequence $1,1,2\ldots$ is formed by adding corresponding terms of the two given sequences. The sum of the first ten terms of the third sequence is:

$\textbf{(A)}\ 978\qquad\textbf{(B)}\ 557\qquad\textbf{(C)}\ 467\qquad\textbf{(D)}\ 1068\\ \textbf{(E)}\ \text{not possible to determine from the information given}$

Solution

Let our geometric sequence be $a,ar,ar^2$ and let our arithmetic sequence be $0,d,2d$. We know that \[\begin{cases} a+0=1 \\ ar+d=1\\ ar^2+2d=2\end{cases}\] This implies that $a=1$, hence $r+d=1$ and $r^2+2d=2$. We can rewrite $r+d=1$ as $d = 1-r$, plugging this into $r^2+2d=2$, we get $r^2+2-2r = 2$, we can simplify this to get $r(r-2)=0$, so $r=0$ or $2$. But since $r\neq 0$, $r=2$ and $d=-1$. So our two sequences are $\{1-n\}$ and $\{2^{n-1}\}$, which means the third sequence will be \[\{2^{n-1}-n+1\}=\{1,1,2,5,12,27,\dots\}\]Calculating the sum of the first 10 terms and adding them up yields 978, hence our answer is $\fbox{A}$.

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