The sum of the first 30 positive integers is 465. Which of the following is the sum of the first 60 positive integers?
A. 465^2
B. 930
C. 1,395
D. 1,830
E. 12,865
The answer is D. 1,830. What formula do I need to know to solve this and is there any way to solve it without a formula?
No formula:
Just add the numbers:
465+31+32+33+34+…+59+60 = 1830.
Using a calculator, 45-60 seconds.
Even if you misread the question and add starting with 1:
1+2+3+…+60 = 1830.
Prob 60-75 seconds.
No need to memorize any formula.
^Don’t listen to that
the formula is fairly simple but you do not need it if you have a graphing calculator you can just set up a summation.
btw the formula is: (Number of terms/2)(first term+Last term)
(60/20)(1+60)=1830
Well the OP did ask for a way to solve it without a formula.
But your formula is very useful to know.
It would be nice to remember every single formula related to Math on the ACT. But if memory fails, keep it simple.
Thanks for the answers! It’s good to know the formula but knowing how to solve it without prior knowledge could be helpful too (:
You can reason it out. If the first 30 numbers sum to 465, then the next 30 numbers will sum to 465 plus an extra 30 for each of the 30 numbers (because they are each 30 larger, right?). So numbers 31-60 will total to 465 plus 30x30 which is 1365. If numbers 1-30 total to 465 and numbers 31-60 total to 1365, then numbers 1-60 total to 1830.
Or you can use the formula above for sum of an arithmetic sequence, if you can remember it LOL.
@chelsalina I only memorize the sum of the first n positive integers (which is n(n+1)/2). The sum of the first n odd (or even) integers is also fairly easy to derive.
Using this formula, the sum of the first 60 positive integers is 60*61/2 = 1830. This should take <15 seconds to multiply out (with or without a calculator).
If you don’t remember the formula (or wish to derive it), here is one way - Define S as follows:
S = 1 + 2 + 3 + … + 60
Then S = 60 + 59 + 58 + … + 1. Write out the two sums side-by-side and add them:
S = 1 + 2 + 3 + … + 60
S = 60 + 59 + … + 1
2S = 61 + 61 + … + 61 = 61*60
Then S = 61*60/2 = 1830.
Another tidbit: the sum of the first n squares (e.g. 1^2 + 2^2 + … + n^2) is fairly well known and is equal to n(n+1)(2n+1)/6. But deriving this is a little trickier.