An arithmetic progression consists of 37 terms. The sum of the first 3 terms of it is 12 and the sum of its last-3 terms is 318, then find the first and last terms
of the progression.​

Answers:

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,

and so on, until,

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,

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.

my answer is$ 12

step-by-step explanation:

because 15 -2=12