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Difference between revisions of "1965 AHSME Problems/Problem 20"
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== Solution ==
 
== 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>.
+
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{\textbf{(C) }6r-1}</math>.
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== See Also ==
 +
{{AHSME 40p box|year=1965|num-b=19|num-a=21}}
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{{MAA Notice}}

Latest revision as of 17:13, 18 July 2024

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 $\boxed{\textbf{(C) }6r-1}$.

See Also

1965 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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