Answer:
The sequence has first term 7 and common difference is 8.
So the sequence is f(n)=7 + 8(n-1)
Step-by-step 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).
