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Difference between revisions of "1965 AHSME Problems/Problem 20"

(Created page with "In order to calculate the <math>r</math>th term of this arithmetic sequence, we can subtract the sum of the first <math>r-1</math> terms from the sum of the first <math>r</mat...")
 
 
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In order to calculate the <math>r</math>th term of this arithmetic sequence, we can subtract the sum of the first <math>r-1</math> terms from the sum of the first <math>r</math> terms of the sequence. Plugging in <math>r</math> and <math>r-1</math> as values of <math>n</math> in the given expression and subtracting yields <cmath>(3r^2+2r)-(3r^2-4r+1)</cmath>. Simplifying gives us the final answer of <math>\boxed{6r-1}</math>.
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== Problem ==
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For every <math>n</math> the sum of n terms of an arithmetic progression is <math>2n + 3n^2</math>. The <math>r</math>th term is:
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<math>\textbf{(A)}\ 3r^2 \qquad \textbf{(B) }\ 3r^2 + 2r \qquad \textbf{(C) }\ 6r - 1 \qquad \textbf{(D) }\ 5r + 5 \qquad \textbf{(E) }\ 6r+2\qquad </math>
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== Solution ==
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In order to calculate the <math>r</math>th term of this arithmetic sequence, we can subtract the sum of the first <math>r-1</math> terms from the sum of the first <math>r</math> terms of the sequence. Plugging in <math>r</math> and <math>r-1</math> as values of <math>n</math> in the given expression and subtracting yields <cmath>(3r^2+2r)-(3r^2-4r+1).</cmath> Simplifying gives us the final answer of <math>\boxed{6r-1}</math>.

Latest revision as of 14:58, 23 June 2016

Problem

For every $n$ the sum of n terms of an arithmetic progression is $2n + 3n^2$. The $r$th term is:

$\textbf{(A)}\ 3r^2 \qquad \textbf{(B) }\ 3r^2 + 2r \qquad \textbf{(C) }\ 6r - 1 \qquad \textbf{(D) }\ 5r + 5 \qquad \textbf{(E) }\ 6r+2\qquad$

Solution

In order to calculate the $r$th term of this arithmetic sequence, we can subtract the sum of the first $r-1$ terms from the sum of the first $r$ terms of the sequence. Plugging in $r$ and $r-1$ as values of $n$ in the given expression and subtracting yields \[(3r^2+2r)-(3r^2-4r+1).\] Simplifying gives us the final answer of $\boxed{6r-1}$.

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