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> | 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> | ||
− | This implies that <math>a=1</math>, hence <math>r+d=1</math> and <math>r^2+2d=2</math>. | + | 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, 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\}< | + | 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}$. |
Revision as of 16:21, 2 July 2020
Problem 45
Given a geometric sequence with the first term and and an arithmetic sequence with the first term . A third sequence is formed by adding corresponding terms of the two given sequences. The sum of the first ten terms of the third sequence is:
Solution
Let our geometric sequence be and let our arithmetic sequence be . We know that This implies that , hence and . We can rewrite as plugging this into we get r(r-2)=0r=02r\neq 0r=2d=-1\{1-n\}\{2^{n-1}\}\fbox{A}$.