If the first term in an arithmetic sequence is 3, and the last term is 136, and the sum is 1390, what are the first 3 terms?
Do you know how many terms there are?
The first three terms are 3, 10, and 17. Since it’s arithmetic, it’s basic addition. If you add up 3, 10, 17, … all the way up to 136, you will get 1390.
but how did you figure that out… 
I mean how do you know the difference between terms?
Okay, this comes sort of naturally for me, but I’ll try to lay it out in clear terms. So since this is arithmetic, the terms’ increase is constant. In this case, we pretty much have to find multiples of the last terms and start guessing (or at least that is what I did). So we know that 3 is the first term, not 0. Therefore, we must subtract 3 from 136 to start getting multiples that could fit our arithmetic equation. We are basically pretending 3 is zero which allows us to evaluate 136. So we take 136 minus 3 and get 133. We start to search for multiples of 133…2(no), 3(no), 4(no), 5(no), 6(no), 7(BINGO!). So if we add 7 to each term starting with three, we will eventually get to 136. All of these numbers will add up to 1390.
ok, thanks…But is there a way to do it if you know the formulas for the Arithmetic sum?
First find the average term by doing this:
(First Term+Last Term)/2= Average Term.
This translates into: (3+136)/2=69.5
Then you need to find how many terms there are in the sequence by doing this:
Sum of terms/Average Term = Number of Terms
This translates into: 1390/69.5=20
Then you do this to find your arithmetic sequence number:
(Last Term - First Term) / (Number of Terms - 1) = Arithmetic Sequence Number
(136-3)/(20-1)=7
Once you have the arithmetic sequence number, it’s pretty much solved. These equations should work for all problems that give you the first number, last number, and sum of the numbers.
A similar approach, but based on another useful fact. If you consider a median (which is also a mean) of the arithmetic sequence set a point of symmetry of that set on the number line, then the sums of each pair of symmetric terms is constant (double of the mean)
For example, in {3,7,11,15,19,23}:
median=mean=(3+23)/2=13
3+23=7+19=11+15=2x13.
.
3+136=139,
there are 1390/139=10 pairs of symmetric numbers in the set, altogether 20 numbers;
The “gap” (common difference) between two adjacent numbers is (136-3)/19=7 because number of gaps is always one less than the number of terms (visualize them on the number line).
3+7=10,
10+7=17.
Yes, there is. Which formulas do you have in mind?
(You can use notation a_n for a sub n.)
@novelidea I like @gcf101’s solution the best.
Here is another way to do it using the general formula for the sum of terms of an arithmetic sequence, but it’s not as pretty.
If the n terms of the arithmetic sequence are a, a+d, a+2d, …, a+(n-1)d, then we have
1390 = an + (n-1)nd/2
a = 3
a + (n-1)d = 136
So (n-1)d = 133, substituting into the original equation we have:
1390 = 3n + 133n/2 or n = 20.
Then use the same reasoning that gcf101 did.
@MITer94
That same formula can be applied in a “shrunken” form:
Sum of n consecutive terms of arithmetic sequence
Sn = [(a1 + a_n) / 2] x n.
1390 = [(a1 + an) / 2] x n
1390 = [(3 + 136) / 2] x n
(notice that (3+136) is one of those “symmetric” sums, and (3+136)/2=median=mean).
n = 1390 / 69.5 = 20, or alternately
n = (1390 x 2) / 139 = (1390/139) x 2 = 20,
and so on.
There is a direct connection, of course, between this algebraic solution and both @adamfromiowa 's and my approaches.
@gcf101 I know; I was just referring to the formula in terms of a and d, which (from what I’ve seen) seems to be more widely taught. Which is why I said that it wasn’t as pretty as previous solutions.
@MITer94 Right, and I wish an alternate formula would be widely taught instead. That’s why I asked @novelidea which formulas he/she had referred to.
@gcf101 Agreed. Gauss would probably be proud.
@MITer94 Well, I don’t think though he would do too hot on the new SAT.
PERFECT! all are so awesome. This is real ACt problem by the way, so hopefully this will help others…
There is always more than one way to skin the act. 
A solution that is based on an intuitive approach is often the easiest.
@adamfromiowa 's one in post #5 is exactly of that kind; it just deserves a little polishing.
The distance (on the number line) between the first and last terms is 136-3=133. Since the “gap” d between the adjacent terms (a common difference) is constant, the number of gaps is 133/d. Assuming that d is an integer **, it should divide 133.
133 has only two factors: 7 and 19.
If d=19, there are 8 terms in the sequence:
3,22,41, … ,136. It’s pretty easy to see that they won’t add up to 1390.
Therefore, d=7, and the first three terms (out of 20) are 3,10,17.
Look Ma, no formulas! 
** In general, d does not have to be an integer.
Or you can just Guess and Check the answers starting with “C”.
As for Gauss, famous story. When he was 5, he was given the problem off adding the numbers from 1 to 100. The tutor thought it would take him several minutes. He immediately said 5050. ( (1 + 100) / 2) * 100. Average the first and last and multiply by the number of elements. This is a useful SAT/ACT skill.