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mathWhat is the common difference of the sum of n terms of an arithmetic series If s=...

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colorcolour | Student, College Freshman | (Level 1) Honors

Posted May 5, 2011 at 2:14 AM via web

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What is the common difference of the sum of n terms of an arithmetic series If s= 5n^2-11n?

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted May 5, 2011 at 6:22 AM (Answer #2)

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To determine the common difference, we'll have to consider to consecutive terms of the a.p.

an - an-1 = d

But s= a1 + a2 + ... + an = 5n^2 - 11n

We'll get the general term an by subtracting both sides the sum: a1 + a2 + .... + an-1:

an = 5n^2 - 11n - (a1 + a2 + .... + an-1)

But a1 + a2 + .... + an-1 = 5(n-1)^2 - 11(n-1)

an = 5n^2 - 11n - 5(n-1)^2 + 11(n-1)

We'll expand the squares:

an = 5n^2 - 11n - 5n^2 + 10n - 5 + 11n - 11

We'll combine and eliminate like terms:

an  = 10n - 16

If we know the expression of the general term an, we can get any term of the arithmetical series.

a1 = 10*1 - 16

a1 = 10 - 16

a1 = -6

a2 = 10*2 - 16

a2 = 20 - 16

a2 = 4

a3 = 10*3 - 16

a3 = 14

The common difference is the difference between 2 consecutive terms:

a2 - a1 = 4 + 6 = 10

d = 10

We can verify and we'll get a3 = a2 + d

14 = 4 + 10

14 = 14

Therefore, the common difference of the given arithmetic series is d = 10.

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted May 7, 2011 at 6:43 AM (Answer #3)

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The sum of n terms of an arithmetic series is given as:

Sn = 5n^2 - 11n

Sn-1 = 5(n - 1)^2 - 11(n - 1)

=> 5(n^2 + 1 - 2n) - 11n + 11

=> 5n^2 + 5 - 10n - 11n + 11

=> 5n^2 - 21n + 16

The term tn = Sn - Sn-1

=> 5n^2 - 11n - 5n^2 + 21n - 16

=> 10n - 16

The common difference = tn - tn-1

=> 10n - 16 - (10(n - 1) - 16)

=> 10n - 16 - 10n + 10 + 16

=> 10

The common difference of the series is 10

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