Can you solve this arithmetic problem

"In an arithmetic sequences, the 1st term is 13 and the 15th term is 111. Find the common difference and the sum of the first 20 terms?"

I honestly don't know how to solve this question, taken from our mathbook.
The answer is "7, 1590"

Could you please solve it and show how you did?

sol'n:

let:
a1 = 1st term
a15 = 15th term
S = sum of the series
n = no. of terms
d = common difference
aN = nth term

aN = a1 + (n - 1)d

let's determine the "d" first:
a15 = a1 + (n - 1)d
111 = 13 + (15 - 1)d
111 = 13 + 15d - d
98 = 14d
d = 7

we can now det. the sum of the series, for 1st 20 terms let's use n = 20:
S = (n/2) * [2a1 + (n - 1)d]
S = (20/2) * [2(13) + (20 - 1)7]
S = 10 * (26 + 133)
S = 1590

Answer: S = 1590 ; d = 7

If the first term is 13, the 15th term is 111, and we're dealing with an arithmetic sequence, let n be the common difference between terms in the sequence. Then the second term in the sequence is
13 + n
The third term is
13 + 2n
and the fourth term is
13 + 3n
From this you can get the pattern I'm trying to hint at. The 15th term in the sequence is
13 + 14n = 111
So 14n = 98
or n = 7
The sum of the first 20 terms?
For n, it's going to be
13 + (13+n) + (13+2n) + ... + (13+19n)
= 20*13 + n(1 + 2 + ... + 19)
= 260 + n((19*20)/2)
Recall that n = 7, so we get
260 + 7*(19*20/2)
= 260 + 7*190
= 260 + 1330
= 1590

accordint to arithmetic sequences recipe:

an = a1 + (n - 1)d, here, an= n'th term a1= 1st term, n = no. of term and d= common difference.
so if the 1st term is 13 and 15th term is 111
so an= 111 a1= 13, n = 15.
so, 111=13+(15-1)d
=> 111-13=14d
=> d=98/14
so, d=7
and the recipe of sum,
Sn=n/2 * [ 2a1 + (n-1)d] where sn= sum of nth term.