Question

The sum of 'n' terms of an arithmetic sequence is 4n^2+3n. What is the first term, the common difference, and the sequence?

Answer 1

The sequence has first term 7 and common difference is 8.

So the sequence is f(n)=7 + 8(n-1)

Explanation:

Let a be the first term.

Let a+d be the second term where d is the common difference.

Then a+2d is the third....

And a+(n-1)d is the nth term.

Adding these terms we get:

an+(n-1)(n)/2×d

For the first term of this sum I seen we had n amount of a's and for the second term I used the well known identity sum of the first n positive integers is n(n+1)/2.

Let's simplify:

an+(n-1)(n)/2×d

Distribute:

an+(n^2d/2)-(nd/2)

Find common denominator:

(2an/2)+(n^2d/2)-(nd/2)

Combine terms into one:

(2an+n^2d-nd)/2

Reorder terms:

(n^2d+2an-nd)/2

Regroup terms:

(n^2d+(2a-d)n)/2

We want the following sum though:

4n^2+3n

This means d/2=4 (so d=8) and (2a-d)/2=3.

So plug d=8 into second equation to solve for a.

(2a-8)/2=3

2a-8=6

2a=14

a=7

The sequence has first term 7 and common difference is 8.

So the sequence is f(n)=7 + 8(n-1).