# Can you solve this arithmetic problem?

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“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.

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

• The nth term of an arithmetic series, Tn is given by Tn = a + (n-1)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 + (n-1)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 + (n-1)d where n happens to be 20.

So a + l = a + a + (n-1)d which is ofcourse 2a + (n-1)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 = 20/2 * (13*2 + (20-1)*7)

S = 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+(n-1)d

We have:

a = 13, n = 15 and a+(n-1)d = 111 Hence:

13+(15-1)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 + (n-1)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+(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.

so here, a1= 13, n = 20, d=7.

so, Sn= 20/2 * [ 2*13 + (20-1)*7]

=1590,

thank you

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