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

See Also

1961 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 25
Followed by
Problem 27
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