In a certain sequence, the first term is 1, and each...

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In a certain sequence, the first term is 1, and each successive term is 1 more than the reciprocal of the term that immediately precedes it. What is the fifth term of the sequence?

(A) 3/5
(B) 5/8
(C) 8/5
(D) 5/3
(E) 9/2

The OA is C.

Is there a strategic approach to this question? Can any experts help, please? Thanks!

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by EconomistGMATTutor » Sun Feb 11, 2018 12:10 pm
In a certain sequence, the first term is 1, and each successive term is 1 more than the reciprocal of the term that immediately precedes it. What is the fifth term of the sequence?

(A) 3/5
(B) 5/8
(C) 8/5
(D) 5/3
(E) 9/2

The OA is C.

Is there a strategic approach to this question? Can any experts help, please? Thanks!
Hi AAPL,
Let's take a look at your question.
$$First\ term\ =\ 1$$
Each successive term is 1 more than the reciprocal of the term that immediately precedes it, therefore, we can find the next term as:
Reciprocal of 1 is 1.
$$Second\ term=\ 1+\frac{1}{1}=1+1=2$$
Reciprocal of 2 is 1/2.
$$Third\ term=\ 1+\frac{1}{2}=\frac{2+1}{2}=\frac{3}{2}$$
Reciprocal of 3/2 is 2/3.
$$Fourth\ term=\ 1+\frac{2}{3}=\frac{3+2}{3}=\frac{5}{3}$$
Reciprocal of 5/3 is 3/5.
$$Fifth\ term=\ 1+\frac{3}{5}=\frac{5+3}{5}=\frac{8}{5}$$

Therefore, option C is correct.

Hope it helps.
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by Scott@TargetTestPrep » Tue Feb 13, 2018 5:37 pm
AAPL wrote:In a certain sequence, the first term is 1, and each successive term is 1 more than the reciprocal of the term that immediately precedes it. What is the fifth term of the sequence?

(A) 3/5
(B) 5/8
(C) 8/5
(D) 5/3
(E) 9/2
First term: 1

Second term: 1 + 1/1 = 2

Third term: 1 + 1/2 = 3/2

Fourth term: 1 + 1/(3/2) = 1 + 2/3 = 5/3

Fifth term: 1 + 1/(5/3) = 1 + 3/5 = 8/5

Answer: C

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