Difference between revisions of "1961 AHSME Problems/Problem 26"
Rockmanex3 (talk | contribs) (Solution to Problem 26) |
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\textbf{(E)}\ 3.5 </math> | \textbf{(E)}\ 3.5 </math> | ||
| − | ==Solution== | + | ==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>. | ||
| − | The <math>50^{\text{th}}</math> term of the sequence is <math>a+49d</math>, so the sum of the first <math>50</math> terms is <math>\frac{50(a + a + 49d}{2}</math>. | + | The <math>50^{\text{th}}</math> term of the sequence is <math>a+49d</math>, so the sum of the first <math>50</math> terms is <math>\frac{50(a + a + 49d)}{2}</math>. |
The <math>51^{\text{th}}</math> term of the sequence is <math>a+50d</math> and the <math>100^{\text{th}}</math> term of the sequence is <math>a+99d</math>, so the sum of the next <math>50</math> terms is <math>\frac{50(a+50d+a+99d)}{2}</math>. | The <math>51^{\text{th}}</math> term of the sequence is <math>a+50d</math> and the <math>100^{\text{th}}</math> term of the sequence is <math>a+99d</math>, so the sum of the next <math>50</math> terms is <math>\frac{50(a+50d+a+99d)}{2}</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. | ||
| + | |||
| + | Theory: | ||
| + | In an AP, we can write | ||
| + | <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: | ||
| + | |||
| + | <math>S_{50} = 2500A+50B=200</math> | ||
| + | |||
| + | <math>S_{100} - S_{50} = 7500A+50B=2700</math> | ||
| + | |||
| + | <math>\implies A=0.5 \text{ and } B = -21</math> | ||
| + | |||
| + | <math>\implies \text{first term} = A+B = \boxed{\textbf{ -20.5 (C) }}</math> | ||
| + | |||
| + | ~Raghav Mukhija(AOPS - CatalanThinker) | ||
==See Also== | ==See Also== | ||
Latest revision as of 09:49, 29 July 2025
Contents
Problem
For a given arithmetic series the sum of the first 
terms is 
, and the sum of the next terms is 
.
The first term in the series is:
Solution 1
Let the first term of the arithmetic sequence be 
and the common difference be 
.
The 
term of the sequence is 
, so the sum of the first terms is 
.
The 
term of the sequence is 
and the 
term of the sequence is 
, so the sum of the next terms is 
.
Substituting in values results in this system of equations.
![\[2a+49d=8\]](http://latex.artofproblemsolving.com/0/f/0/0f0a22c9b30a18794b65bac55dedea966df6b8e3.png)
![\[2a+149d=108\]](http://latex.artofproblemsolving.com/3/e/c/3ec72eaa5bd65899b44ed0dddf9b6260ec7619a8.png)
Solving for yields 
. The first term is 
, which is answer choice 
.
Solution 2
Same as solution 1 but faster.
Theory:
In an AP, we can write
, where common difference(
and first term(
This is already known, and not part of the solution :) .
So, given:
~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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
