How do you do this problem?

Question

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?

Answer 1

an = a1 + (n - 1)d, where

a1 is the first term of the arithmetic progression
an is the nth term of the arithmetic progression
n is the term number
d is the common difference of the arithmetic progression

136 = 3 + (n - 1). d


Sum of n terms of an AP is given by

Sn = n ( a1 + an ) / 2

1390 = n (3 + 136) / 2 or n = 2X1390 / 139 = 20

So, 136 = 3 + 19.d or 19d = 136 - 3 = 133

And d = 133/19 = 7

So, the first 3 terms will be 3, 10 and 17

Answer 2

Ooh, I love these problems! OK, here's one approach: since it's an arithmetic series the numbers are 3, 3+n, 3+2n and so on. The value of n has to be such that, adding it up a bunch of times, you'll end up at exactly 136. That is, n has to be a divisor of 136-3 = 133. Now 133 = 7 * 19. So n has to be either 7 or 19. I don't have my calculator handy, but I'm guessing that the answer is 19 because, if the answer were 7, you'd be adding up so many terms that you'd get past 1,390 in a hurry.

[edit]
Whoops, I was wrong! It's 7, not 19; I just did the math. So the first three terms are 3,10,17. Sorry!

Answer 3

PA.
a=3,S=1390
S=n/2[a+l]
1390=n/2[3+l]
n/2=1390/139=10
n=20
l=a+(n-1)d
136=3+19d
133=19d
d=133/19=7
the first three terms are 3,10,17 answer
verify?a(20)=3+19*7=136
S=20/2[3+136]=10*139=1390

Answer 4

We know that
1390=3+?+?+...+136
so
1390=136+...+?+?+3
so
2780=139+139+139+...+139 (n times)
n=2780/139=20
so the constant difference a is given by
3+19*a=136
so
a=7
thus first three terms are 3,10,17.

Answer 5

Sn = (n/2) * (a0 + an)
a0 = 3
an = 136
Sn = 1390
So we can get n = 20

Now an = a0 + (n-1)*d
136 = 3 + 19d
d = 7
So
a0 = 3,
a1 = 10
a2 = 17

Answer 6

3,10,17