Answers:
- First term = 1
- Last term (37th term) = 109
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Step-by-step explanation:
a1 = first term
a2 = second term
a3 = third term
and so on, until we reach a37 to represent the 37th term
We don't know the first term, but we'll call it x for now. We'll also let d be the difference between each adjacent term, aka the common difference.
Meaning that,
- a1 = x
- a2 = a1+d = x+d
- a3 = a2+d = (x+d)+d = x+2d
- a4 = a3+d = (x+2d)+d = x+3d
and so on, until,
- a35 = x+34d
- a36 = x+35d
- a37 = x+36d
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Add up the first three terms (a1 through a3) and set the sum equal to 12
Then do a bit of algebra to get
a1+a2+a3 = 12
(x)+(x+d)+(x+2d) = 12
3x+3d = 12
3(x+d) = 12
x+d = 12/3
x+d = 4
x = 4-d
Do the same for the last three terms, but the sum here is 318
a35+a36+a37 = 318
(x+34d)+(x+35d)+(x+36d) = 318
3x+105d = 318
3(x) + 105d = 318
3(4-d)+105d = 318 .... replace x with 4-d; isolate d
12-3d+105d = 318
102d+12 = 318
102d = 318-12
102d = 306
d = 306/102
d = 3
The common difference is 3. It means each term is increasing by 3.
Use this value of d to find x
x = 4-d
x = 4-3
x = 1
The first term is 1. That makes the second term x+d = 1+3 = 4, then the third term is x+2d = 1+2*3 = 7, and so on.
The sum of the first three terms is 1+4+7 = 12. That helps partly confirm the answer.
Also we have,
- a35 = x+34d = 1+34(3) = 103
- a36 = x+35d = 1+35(3) = 106
- a37 = x+36d = 1+36(3) = 109 is the last term (37th term)
Or you could note that the 35th term is 103, and then you increase by 3 each time to get the remaining two terms.
The last part of the confirmation is adding up those three last terms: 103+106+109 = 318, so the answer is fully verified at this point.
Side note: An arithmetic sequence never ends. There technically is no "last term"; however, since we're confined to focusing on these specific 37 terms, we can get away with using the phrasing.