# 1961 AHSME Problems/Problem 26

## Problem

For a given arithmetic series the sum of the first $50$ terms is $200$, and the sum of the next $50$ terms is $2700$. The first term in the series is: $\textbf{(A)}\ -1221 \qquad \textbf{(B)}\ -21.5 \qquad \textbf{(C)}\ -20.5 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 3.5$

## Solution

Let the first term of the arithmetic sequence be $a$ and the common difference be $d$.

The $50^{\text{th}}$ term of the sequence is $a+49d$, so the sum of the first $50$ terms is $\frac{50(a + a + 49d)}{2}$.

The $51^{\text{th}}$ term of the sequence is $a+50d$ and the $100^{\text{th}}$ term of the sequence is $a+99d$, so the sum of the next $50$ terms is $\frac{50(a+50d+a+99d)}{2}$.

Substituting in values results in this system of equations. $$2a+49d=8$$ $$2a+149d=108$$ Solving for $a$ yields $a = \frac{-41}{2}$. The first term is $-20.5$, which is answer choice $\boxed{\textbf{(C)}}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 