Difference between revisions of "1955 AHSME Problems/Problem 45"
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− | 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= | + | 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> |
− | This implies that <math>a=1</math>, hence <math>r+d=1</math> and <math>r^2+2d= | + | This implies that <math>a=1</math>, hence <math>r+d=1</math> and <math>r^2+2d=2</math>. Solving this system yields <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 16:19, 2 July 2020
Problem 45
Given a geometric sequence with the first term and and an arithmetic sequence with the first term . A third sequence is formed by adding corresponding terms of the two given sequences. The sum of the first ten terms of the third sequence is:
Solution
Let our geometric sequence be and let our arithmetic sequence be . We know that This implies that , hence and . Solving this system yields , so or . But since , and . So our two sequences are and , which means the third sequence will be Calculating the sum of the first 10 terms and adding them up yields 978, hence our answer is .