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

1961 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 25	Followed by Problem 27
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
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1961 AHSME Problems/Problem 26

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