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 1

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

Solution 2

Same as solution 1 but faster.

Theory: In an AP, we can write $S_n = An^2+Bn$ , where common difference( $d) = 2A$ and first term( $a) = A(1^2)+B(1) = A+B$

This is already known, and not part of the solution :) .

So, given:

$S_{50} = 2500A+50B=200$

$S_{100} - S_{50} = 7500A+50B=2700$

$\implies A=0.5 \text{ and } B = -21$

$\implies \text{first term} = A+B = \boxed{\textbf{ -20.5 (C) }}$

~Raghav Mukhija(AOPS - CatalanThinker)

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

Contents

Problem

Solution 1

Solution 2

See Also