“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?
6 Answers

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

The nth term of an arithmetic series, Tn is given by Tn = a + (n1)d
where a is the first term and d is the common difference.
I’m going to assume you know what a ‘common difference’ is, generally what an arithmetic series is
and that you don’t need proof of any general formula like the one above.
If the 1st term is 13 then a = 13
If the 15th term is 111 then
Tn = a + (n 1 )d
111 = 13 + (15 1)d –> Note the substitutions of Tn = 111 , a = 13 and n = 15
Rearranging and solving for d, d = 98/14 = 7
The common difference is 7.
The sum to n terms of an arithmetic series is given by Sn = n/2 * (2a + (n1)d)
And also in some books by Sn = (n/2)(a + l)
Notice though, that in the first 20 terms, which you are trying to find, the last term is ofcourse the 20th term which is given by l = a + (n1)d where n happens to be 20.
So a + l = a + a + (n1)d which is ofcourse 2a + (n1)d, which saves you the trouble of having to separately find the last term as “l” is now out of the equation and it only has a, n and d in it.
This corresponds to S[20] = 20/2 * (13*2 + (201)*7)
S[20] = 10(26 + 133) = 10(159) = 1590.
The sum to 20 terms is 1590.

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

The n’th term of an arithmetic sequence is: a+(n1)d
We have:
a = 13, n = 15 and a+(n1)d = 111 Hence:
13+(151)d = 111
=> 14d = 98
=> d = 7
Ans: Common difference d = 7
The formula for the sum Sn of arithmetic series is:
Sn = n/2{2a + (n1)d] Hence:
S_20 = 20/2 [2*13 + (20 – 1)7]
=> S_20 = 10[26 + 133]
=> S_20 = 1590
Ans: The sum of the first 20 terms = 1590

an arithmetic sequence is a sequence of which the terms increase with a fixed amount. This fixed amount is called the common difference, let’s denote it with r for now.
This way, the sequence would be as follows:
13 + 0r
13 + 1r
13 + 3r
…
13 + 14 r
From the fact that the 15th term is 111, we can find the value r:
13 + 14 r = 111
and thus: 14 r = 111 – 13 or : r = 98 / 14 = 7

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+(151)d
=> 11113=14d
=> d=98/14
so, d=7
and the recipe of sum,
Sn=n/2 * [ 2a1 + (n1)d] where sn= sum of nth term.
so here, a1= 13, n = 20, d=7.
so, Sn= 20/2 * [ 2*13 + (201)*7]
=1590,
thank you