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

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

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11 Apr 2019, 07:30
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00:00

Difficulty:

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

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11 Apr 2019, 07:59
1
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.
Attachments

Scan.pdf [530.77 KiB]

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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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