1960 AHSME Problems/Problem 36: Difference between revisions

Revision as of 18:20, 17 May 2018

Problem

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 $\boxed{\textbf{(B)}}$ .

1960 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 35	Followed by Problem 37
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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 ==See Also==
 {{AHSME 40p box|year=1960|num-b=35|num-a=37}}
+[[Category:Introductory Algebra Problems]]

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1960 AHSME Problems/Problem 36: Difference between revisions

Revision as of 18:20, 17 May 2018

Problem

Solution

See Also