For the infinite sequence a1, a2, a3, ... an, an+1, an=3(an−1) for all

Question

For the infinite sequence  a_1 ,  a_2 ,  a_3 , ...  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

ScottTargetTestPrep · Accepted Answer

We are given that a(n) = 3 x a(n-1)

Thus:

a(2) = 3 x a(1)

a(3) = 3 x 3 x a(1) = 9a(1)

a(4) = 3 x 9 x a(1) = 27a(1)

a(5) = 3 x 27 x a(1) = 81a(1)

We are given a(5) - a(2) = 156. Thus:

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

78a(1) = 156

a(1) = 2

Answer: B

Kurtosis · Answer

a5 = 3(a4) = 9(a3) = 27(a2) = 81(a1)

a5 - a2 = 81(a1) - 3(a1) = 156
78(a1) = 156
a1 = 2

Answer: B

0akshay0 · Answer

a_n=3*a_{n−1} a_5-a_2 = 156 a_5 = 27*a_2 a_2 = 156/26 = 6 = 3*a_1 a_1 = 2 Hence Option B is correct. Hit Kudos if you liked it 8-)

sisirkant · Answer

Answer is B (a1= 2)

a5-a2=156
=> 3*a4-3*a1=156
=>a4-a1=52 ---- (1)

a4= 3*a3=3*3*a2=3*3*3*a1 ---- (2)

Solving (1) and (2), a1=2

Diwakar003 · Answer

Option A can be eliminated as  a1  is negative so other elements in the series should also be increasingly negative. Hence the sum  a_5−a_2  should also be negative which is not the case here.

Option B) Writing down the elements with  a1=2  , 2,6,18,54,162. So 162-6 = 156.

Hence B

Cheers!

RR88 · Answer

Option B

an = 3*an-1; n>1.

a5-a2 =156

3^4a1 - 3^1a1 = 156

a1 =2

matthewsmith_89 · Answer

An easier way to solve this problem is process elimination.

Let's start with A
If a1 = -1 then the sequences would be: -1, -3, -9, -27, -81
In this case a5 - a2 is not equal 156 so A is out

B
If a1 = 2 then the sequences would be: 2, 6, 18, 54, 162
162 - 5 = 156

Answer is B :-D

maheshaero20 · Answer

The series is in Geometric progression.

\frac{a_5}{a_2}=3^3

solving we get  27a_2-a_2=156  ==>  a_2=6

a_2=3*a_1  ==>  a_1=2

LeoN88 · Answer

a5 = a4 * 3
a4 = a3 * 3...

a5 - a2 = 156 = (81-3)a1, compute to ind a1

OA 2.

bumpbot · Answer

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For the infinite sequence a1, a2, a3, ... an, an+1, an=3(an−1) for all

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For the infinite sequence a1, a2, a3, ... an, an+1, an=3(an−1) for all

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