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The sequence of numbers a, ar, ar^2, and ar^3

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The sequence of numbers a, ar, ar^2, and ar^3  [#permalink]

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New post 11 Apr 2019, 07:30
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A
B
C
D
E

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  35% (medium)

Question Stats:

83% (02:05) correct 17% (02:09) wrong based on 12 sessions

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The sequence of numbers \(a\), \(ar\), \(ar^2\), and \(ar^3\) are in geometric progression. The sum of the first four terms in the series is 5 times the sum of first two terms and \(r\neq{–1}\). How many times larger is the fourth term than the second term?

A) 1

B) 2

C) 4

D) 5

E) 6

Source: GMAT Math Bible p.319

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The sequence of numbers a, ar, ar^2, and ar^3  [#permalink]

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New post 11 Apr 2019, 07:59
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Solution attached.

Edit 13/04/'19: Mahmoudfawzy83 asked me to rewrite the solution as forum post, which I pursued and illustrated in the .pdf attached.

So, we have this sequence: \(a, ar, ar^2, ar^3\).

We need to find how many times larger the fourth term is with respect to the second term. So, we are taking into account \(ar\) and \(ar^3\). How many times... means comparing them to find the ratio. So, \(\frac{ar^3}{ar} = r^2\) represents how many times the fourth term is larger than the second term.

Next step: we know that the sum of first four terms are 5 times the sum of first two terms. We can write:

\(a, ar, ar^2, ar^3 = 5 (a + ar)\)

Arrange it as:

\(a(1 + r + r^2 + r^3) = 5a (r + 1)\)
\(a[1(r^2+1)+r(r^2+1)] = 5a (r + 1)\)
\(a(r+1)(r^2+1) = 5a (r + 1)\)
\(a(r^2+1) = 5a\)
\(r^2 + 1 = 5\)
\(r^2 = 4\)

Hope it's clear.
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The sequence of numbers a, ar, ar^2, and ar^3   [#permalink] 11 Apr 2019, 07:59
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