Common difference.The sum of n terms of an arithmetic series is 5n^2-11n for all values of n. Detemine the common difference.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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The problem provides the information that `sum_(k=1)^n b_k = 5n^2-11n` , hence, evaluating the general term, yields:

`b_n = sum_(k=1)^n b_k - sum_(k=1)^(n-1) b_k`

Replacing `5n^2-11n` for ` sum_(k=1)^n b_k` and `5(n - 1)^2 - 11(n - 1)` for `sum_(k=1)^(n-1) b_k` yields:

`b_n = 5n^2 - 11n - 5(n - 1)^2 + 11(n - 1)`

`b_n = 5n^2 - 11n - 5n^2 + 10n - 5 + 11n - 11`

`b_n = 10n - 11`

You need to evaluate the common difference such that:

`b_2 = b_1 + d => d = b_2 - b_1`

Replacing 1 for n yields:

`b_1 = 10 - 11 => b_1 = -1`

Replacing 2 for n yields:

`b_2 = 20 - 11 => b_2 = 9`

`d = 9 - (-1) => d = 10`

Hence, evaluating the common difference, under the provided conditions, yields d = 10.

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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

We'll get:

an - an-1 = d

We know, from enunciation, that the sum of n terms of the a.p. is:

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

We'll determine 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

Knowing the general term an, we can determine 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

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

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