# 1960 AHSME Problems/Problem 36

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