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The nth term of an increasing sequence S is given by Sn = Sn-1 + Sn-2 [#permalink]

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01 Jun 2017, 21:14

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The nth term of an increasing sequence S is given by \(S_n = S_{n-1} + S_{n-2}\) for n > 2 and the nth term of a sequence S’ is given by \(S’_n = S’_{n-1} - S’_{n-2}\) for n > 2. If \(S_5 = S’_5\), what is the average (arithmetic mean) of \(S_2\) and \(S’_2\)?

(1) The difference between the fourth term and the second term of sequence S is 14.

(2) The sum of the fourth term and the second term of sequence S’ is 14.

Re: The nth term of an increasing sequence S is given by Sn = Sn-1 + Sn-2 [#permalink]

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01 Jun 2017, 22:13

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Let the first 5 terms of sequence S be represented as: S1, S2, S3, S4, S5 Here S3 = S2+S1, S4 = S3+S2, S5 = S4+S3 ..

Let the first 5 terms of sequence S' be represented as: S1', S2', S3', S4', S5' Here S4' = S3'-S2', S5' = S4'-S3'

Now lets look at the statements: (What I will try to do is write as many terms as possible in terms of either S2 or S2' because our objective is to calculate mean of S2 & S2', which we will get once we have the sum S2+S2')

Statement 1. S4-S2 = 14, Or S4 = 14+S2, But S4 = S3+S2.. this means S3 = 14 Thus S5 = S4+S3 = 14+S2 + 14 = 28+S2 So S5 can be written in terms of S2 as '28+S2'.

We are given that S5 = S5', so S5' = 28+S2 Now, S5' = S4' - S3' and S4' = (S3'-S2') - S3' = -S2'

See, S5' can be written as (28+S2) and it can also be written as '-S2' . Equating them both: 28+S2 = -S2' Or S2+S2' = -28 We have their sum, so their average = -28/2. Sufficient.