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Difference between revisions of "1961 AHSME Problems/Problem 26"

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\textbf{(E)}\ 3.5    </math>
 
\textbf{(E)}\ 3.5    </math>
  
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==Solution==
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==Solution 1==
 
Let the first term of the [[arithmetic sequence]] be <math>a</math> and the common difference be <math>d</math>.
 
Let the first term of the [[arithmetic sequence]] be <math>a</math> and the common difference be <math>d</math>.
  
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<cmath>2a+149d=108</cmath>
 
<cmath>2a+149d=108</cmath>
 
Solving for <math>a</math> yields <math>a = \frac{-41}{2}</math>.  The first term is <math>-20.5</math>, which is answer choice <math>\boxed{\textbf{(C)}}</math>.
 
Solving for <math>a</math> yields <math>a = \frac{-41}{2}</math>.  The first term is <math>-20.5</math>, which is answer choice <math>\boxed{\textbf{(C)}}</math>.
 +
 +
==Solution 2==
 +
Same as solution 1 but faster.
 +
 +
We know that
 +
 +
<math>S_n = An^2+Bn</math>, where common difference(<math>d) = 2A</math> and first term(<math>a) = A(1^2)+B(1) = A+B</math>
 +
 +
This is already known, and not part of the solution :) .
 +
 +
So, given:
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 +
<math>S_{50} = 2500A+50B=200</math>
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<math>S_{100} - S_{50} = 7500A+50B=2700</math>
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<math>\implies A=\frac{1}{2}=0.5 \text{ and } B = -21</math>
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<math>\implies \text{first term} = A+B = \boxed{\textbf{ -20.5 (C) }}</math>
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~Raghav Mukhija(AOPS - CatalanThinker)
  
 
==See Also==
 
==See Also==

Revision as of 09:47, 29 July 2025

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.

We know that

$S_n = An^2+Bn$, 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:

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

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

$\implies A=\frac{1}{2}=0.5 \text{ and } B = -21$

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

~Raghav Mukhija(AOPS - CatalanThinker)

See Also

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
All AHSME Problems and Solutions

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