Difference between revisions of "1960 AHSME Problems/Problem 36"

(Solution for Problem 36)
 
m (Solution)
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<cmath>2dn^2</cmath>
 
<cmath>2dn^2</cmath>
  
The answer is <math>\boxed{\textbf{B}}</math>.
+
The answer is <math>\boxed{\textbf{(B)}}</math>.
  
 
==See Also==
 
==See Also==
 
{{AHSME 40p box|year=1960|num-b=35|num-a=37}}
 
{{AHSME 40p box|year=1960|num-b=35|num-a=37}}

Revision as of 00:20, 11 May 2018

Problem 36

Let $s_1, s_2, s_3$ be the respective sums of $n, 2n, 3n$ terms of the same arithmetic progression with $a$ as the first term and $d$ as the common difference. Let $R=s_3-s_2-s_1$. Then $R$ is dependent on:

$\textbf{(A)}\ a\text{ }\text{and}\text{ }d\qquad \textbf{(B)}\ d\text{ }\text{and}\text{ }n\qquad \textbf{(C)}\ a\text{ }\text{and}\text{ }n\qquad \textbf{(D)}\ a, d,\text{ }\text{and}\text{ }n\qquad  \textbf{(E)}\ \text{neither} \text{ } a \text{ } \text{nor} \text{ } d \text{ } \text{nor} \text{ } n$

Solution

The nth term of the sequence with first term $a$ is $a + d(n-1)$. That means the sum of the first $n$ terms is with first term $a$ is $\frac{n(2a + dn -d)}{2}$.

The 2nth term of the sequence with first term $a$ is $a + d(2n-1)$. That means the sum of the first $2n$ terms is with first term $a$ is $\frac{2n(2a + 2dn -d)}{2}$.

The 3nth term of the sequence with first term $a$ is $a + d(3n-1)$. That means the sum of the first $3n$ terms is with first term $a$ is $\frac{3n(2a + 3dn -d)}{2}$.

Substituting $s_1$, $s_2$, and $s_3$ with its respective values results in

\[\frac{3n(2a + 3dn -d)}{2}-\frac{2n(2a + 2dn -d)}{2}-\frac{n(2a + dn -d)}{2}\] \[\frac{6an + 9dn^2 -3nd}{2}-\frac{4an + 4dn^2 - 2nd}{2}-\frac{2an + dn^2 - dn}{2}\]

\[\frac{4dn^2}{2}\]

\[2dn^2\]

The answer is $\boxed{\textbf{(B)}}$.

See Also

1960 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 35
Followed by
Problem 37
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