I was given this question on a college assessment pre-test. I got the correct answer in a reasonable amount of time, but mostly because I worked backwards and double checked my answer. After I was done, I tried to find math on the net to solve it correctly, but I feel there is probably a simpler solution than this.
Is there an easier way to solve problems in this template?
Given the first and n-th values in an arithmetic progression, and the sum of the progression up to n (inclusive), give the first x terms of the series.
The actual question on the quiz
In an arithmetic series, the terms of the series are equally spread out. For example, in 1 + 5 + 9 + 13 + 17, consecutive terms are 4 apart. If the first term in an arithmetic series is 3, the last term is 136, and the sum is 1,390, what are the first 3 terms?
This one lended itself to intuition and backwards-work because I was reasonably certain that d would be an integer, and 133 (
136 - 3) is evenly divisible by 7. But I would like to find a simpler way to solve problems like this in the future.
The work I did to solve the question
I used the formulas:
- Sn = ½ n(a1 + an)
- an = dn + c
Where n is the count of values in the sequence, d is the common difference (distance between values in the sequence), and c is an unknown constant
For the first equation, I solved for n:
1390 = ½ n(3 + 136)
1390 = ½ n(139)
10 = ½ n
20 = n
For the second equation, I first solved for c, by making two equations out of the values I had for an, and the previously calculated value for n:
3 = d + c,
136 = 20d + c
(with the first): d = 3 - c
(substitute): 136 = 20(3 - c) + c
136 = 60 - 19c
76 = -19c
-4 = c
Then I solved for d:
(substitute): 3 = d - 4
d = 7
(double-check): 136 = 20d - 4
140 = 20d
7 = d
The final answer was 3, 3 + d, 3 + 2d:
3, 10, 17
Which I did a brute force double-check on.