Can you solve this arithmetic problem?

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 + (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] = 20/2 * (13*2 + (20-1)*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+(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