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If p is a constant and an−1 + an = pn (n − 1) for all positive integer

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Joined: 02 Sep 2009
Posts: 58335
If p is a constant and an−1 + an = pn (n − 1) for all positive integer  [#permalink]

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New post 26 Sep 2018, 05:28
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Difficulty:

  65% (hard)

Question Stats:

54% (02:33) correct 46% (02:24) wrong based on 52 sessions

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Re: If p is a constant and an−1 + an = pn (n − 1) for all positive integer  [#permalink]

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New post 26 Sep 2018, 14:13
Bunuel wrote:
If p is a constant and \(a_{n−1} + a_n = pn (n − 1)\) for all positive integers n, what is the value of p?


(1) \(a_{31} − a_{29} = 120\)

(2) \(a_2 = 6\)


Since it's not entirely clear where to start, we'll try using a few numbers to help us see the logic.
This is an Alternative approach.

It often helps to start small: let's try n = 1.
This gives a_0 + a_1 = 0 meaning a_1 = -a_0
Trying n = 2 gives a_1 + a_2 = 2p
Substitution gives a_2 = 2p + a_0
Trying n = 3 gives a_2 + a_3 = 6p
Substituing the above equation gives a_3 = 4p - a_0
And we can already see the logic: every value of a_n is determined by the value of p and the value of a_0, which changes signs.
Just to verify: n = 4 gives a_3 + a_4 = 12p so substitution would give a_4 = 8p + a_0, as expected.

Now that we have some intuition for the sequence, we can look at our data.
(1) from our work above we know that a_31 and a_29 have the same sign for a_0 so subtracting them will cancel out a_0 leaving only (some number)*p = 120
We can therefore directly solve for p.
Sufficient.

(2) Again based on the above, this is not sufficient: we can substitute a_2 = 6 into the above equations but will still have one equation with 2 variables left which is unsolvable.
Insufficient.

(A) is our answer
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Re: If p is a constant and an−1 + an = pn (n − 1) for all positive integer   [#permalink] 26 Sep 2018, 14:13
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If p is a constant and an−1 + an = pn (n − 1) for all positive integer

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