For the infinite sequence a_1, a_2, a_3, ... a_n, a_n+1

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For the infinite sequence $$a_1, a_2, a_3, \cdots, a_{n}, a_{n+1}$$ $$a_{n}=3*a_{n-1}$$ for all n>1. If $$a_5 - a_2 = 156,$$ what is $$a_1?$$

A. -1
B. 2
C. 3
D. 4
E. 8

The OA is B.

Please, can anyone help me to solve this PS question? Thanks in advance!

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by [email protected] » Wed May 23, 2018 9:09 pm
Hi All,

With sequence questions, you will almost always be given 'instructions' about how the sequence works - and those instructions will be technical in nature. Here, the wording ultimately means that each 'term' after the first term is equal to 3 TIMES the prior term. For example, a sequence built on this definition COULD be 1, 3, 9, 27, 81, 243, etc. There is an additional piece of information though: we're told that the (5th term) - (2nd term) = 156. We're asked for the value of the 1st term. This question can be solved by TESTing THE ANSWERS.

Based on the initial example I listed, the difference between the 5th and 2nd terms would be 81 - 3 = 78... which is exactly HALF of what we need it to be. This means that the first terms MUST be greater than 1; based on the Answer choices, let's TEST Answer B first....

Answer B: 2
If the first term is 2, then the terms in the sequence would be....
2, 6, 18, 54, 162
The 5th term - the 2nd term = 162 - 6 = 156
This is an exact match for what we were told, so this MUST be the answer.

Final Answer: B

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Rich
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by Scott@TargetTestPrep » Thu May 24, 2018 4:57 pm
BTGmoderatorLU wrote:For the infinite sequence $$a_1, a_2, a_3, \cdots, a_{n}, a_{n+1}$$ $$a_{n}=3*a_{n-1}$$ for all n>1. If $$a_5 - a_2 = 156,$$ what is $$a_1?$$

A. -1
B. 2
C. 3
D. 4
E. 8
We see that:

a(2) = 3a(1)

a(3) = 9a(1)

a(4) = 27a(1)

a(5) = 81a(1)

Thus:

a(5) - a(2) = 156

81a(1) - 3a(1) = 156

78a(1) = 156

a(1) = 2

Answer: B

Scott Woodbury-Stewart
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