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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2508
Given Kudos: 1053
GPA: 3.39
In a geometric sequence, each term is a constant multiple of
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12 Mar 2020, 04:53
12 Mar 2020, 04:53
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In a geometric sequence, each term is a constant multiple of the preceding one. What is the sum of the first four numbers in a geometric sequence whose second number is 4 and whose third number is 6?
(A) 16
(B) 19
(C) 22\(\frac{1}{2}\)
(D) 21\(\frac{2}{3}\)
(E) 20
(A) 16
(B) 19
(C) 22\(\frac{1}{2}\)
(D) 21\(\frac{2}{3}\)
(E) 20
MyGuru Representative
Joined: 09 Apr 2020
Posts: 35
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Re: In a geometric sequence, each term is a constant multiple of
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18 Apr 2020, 20:18
18 Apr 2020, 20:18
1
Expert Reply
GeminiHeat wrote:
In a geometric sequence, each term is a constant multiple of the preceding one. What is the sum of the first four numbers in a geometric sequence whose second number is 4 and whose third number is 6?
(A) 16
(B) 19
(C) 22\(\frac{1}{2}\)
(D) 21\(\frac{2}{3}\)
(E) 20
(A) 16
(B) 19
(C) 22\(\frac{1}{2}\)
(D) 21\(\frac{2}{3}\)
(E) 20
Since the ratio between successive numbers in a geometric sequence is always the same, use the numbers given in the problem to figure out what that ratio is.
\(\frac{6}{4} = 1.5\) so each number in the sequence is 1.5 times the previous number.
The first number is \(\frac{4}{1.5}\) or \(\frac{4}{(3/2)} = 4*\frac{2}{3} = \frac{8}{3}\)
The last number is 6*1.5 = 9
So the sum of all four numbers is: \(\frac{8}{3} + 4 + 6 + 9 = 21\frac{2}{3}\)
Answer: D
gmatclubot
Re: In a geometric sequence, each term is a constant multiple of [#permalink]
18 Apr 2020, 20:18
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