Sequences

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Sequences

by DJHJA » Sun Dec 19, 2010 11:01 pm
In the infinite sequence, a1, a2, a3........a(n) each term after the first is equal to twice the previous term,

if a5 - a2=12, what is the value of a1?


a. 4
b. 24/7
c. 2
d. 12/7
e. 6/7


ans is 6/7.. this is a pretty easy question, not sure y i cant get it :( can anyone help?

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by Rahul@gurome » Sun Dec 19, 2010 11:08 pm
DJHJA wrote:In the infinite sequence, a1, a2, a3........a(n) each term after the first is equal to twice the previous term, if a5 - a2=12, what is the value of a1?
  • a. 4
    b. 24/7
    c. 2
    d. 12/7
    e. 6/7
Clearly the terms are in geometric progression with first term a1 and common ratio 2.

Say a(1) = a, then the n-th term is given by a(n) = 2^(n - 1)*a

Now a(5) - a(2) = 12
=> [2^(5 - 1)*a] - [2^(2 - 1)*a] = 12
=> 16a - 2a = 12
=> 14a = 12
=> a = 6/7

The correct answer is E.
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by DJHJA » Mon Dec 20, 2010 12:37 pm
[2^(5 - 1)*a] - [2^(2 - 1)*a] = 12
= 16a - 2a = 12


2^ means squared, correct?

if so then 5-1 = 4, 4^2= 16*a= 16a,


and 2-1=1, 1^2=1*a=1a,


so shouldnt it be 16a-1a=12? not sure if im following

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by GMATGuruNY » Mon Dec 20, 2010 12:56 pm
DJHJA wrote:In the infinite sequence, a1, a2, a3........a(n) each term after the first is equal to twice the previous term,

if a5 - a2=12, what is the value of a1?


a. 4
b. 24/7
c. 2
d. 12/7
e. 6/7


ans is 6/7.. this is a pretty easy question, not sure y i cant get it :( can anyone help?

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To get from a2 to a5, we need to multiply by 2 three times.
Thus, a5 = 2*2*2*a2 = 8a2.

Substituting a5 = 8a2 into a5 - a2 = 12, we get:
8a2 - a2 = 12.
7a2 = 12.
a2 = 12/7.

Since a2 = 2a1, we get:
12/7 = 2a1
a1 = 6/7.

The correct answer is E.
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by Rahul@gurome » Mon Dec 20, 2010 1:04 pm
DJHJA wrote:2^ means squared, correct?

if so then 5-1 = 4, 4^2= 16*a= 16a,

and 2-1=1, 1^2=1*a=1a,


so shouldnt it be 16a-1a=12? not sure if im following
No.
2^(x) means 2 to the power x = 2 is multiplied with itself x times
Whereas (x)^2 means x square = x is multiplied with itself twice

Thus, 2^(5 - 1)a = (2^4)a = 2*2*2*2*a = 16a
and 2^(2 - 1)a = (2^1)a = 2a
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by showbiz » Mon Dec 20, 2010 5:43 pm
There is a simple way of doing it, but I have seen that you need a little practice to get some heart to do it.

a1=a1
a2=2a1
a3=2a2=4a1
a4=2a3=8a1
a5=2a4=16a1

16a1-2a1=12
14a1=12
a1=12/14=6/7