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# The sequence a(1), a(2), ..., a(n), ... is such that an=4a(n-1} - 3

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Re: The sequence a(1), a(2), ..., a(n), ... is such that an=4a(n-1} - 3 [#permalink]
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE
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Re: The sequence a(1), a(2), ..., a(n), ... is such that an=4a(n-1} - 3 [#permalink]
The sequence a(1), a(2), …, a(n), … is such that a(n)=4a(n–1) –3 for all integers n>1. If a(3)=x, then a(1)=?

A. 4x–3
B. 16x–15
C. (x+3)/4
D. (x+3)/16
E. (x+15)/16

I love such questions!

Since we know a(3) we can find a(2), if we can find a(2) we can find a(1). a(3)=4a(2)-3 ---> x=4a(2)-3 ---> a(2)=(x+3)/4

a(2)=4a(1)-3 ---> (x+3)/4=4a(1)-3 ---> 4a(1)=(x+15)/4 ---> a(1)=(x+15)/16 The answer is E.
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Re: The sequence a(1), a(2), ..., a(n), ... is such that an=4a(n-1} - 3 [#permalink]
1
Kudos
a(1)= {a(2)+3}/4
Splitting the denominator a(1)= a(2)/4 +3/4 say eq----i
a(2)={a(3)+3}/4
a(2)={x+3}/4
a(2)/4= {x+3}/16 let say this is eq -----ii

Sub a(2)/4 in eq i

a(1)= {x+3}/16 + 3/4
a(1)= (x+15)/16
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Re: The sequence a(1), a(2), ..., a(n), ... is such that an=4a(n-1} - 3 [#permalink]
ziko
The sequence a(1), a(2), …, a(n), … is such that a(n)=4a(n–1) –3 for all integers n>1. If a(3)=x, then a(1)=?

A. 4x–3
B. 16x–15
C. (x+3)/4
D. (x+3)/16
E. (x+15)/16

I love such questions!

Since we know a(3) we can find a(2), if we can find a(2) we can find a(1). a(3)=4a(2)-3 ---> x=4a(2)-3 ---> a(2)=(x+3)/4

a(2)=4a(1)-3 ---> (x+3)/4=4a(1)-3 ---> 4a(1)=(x+15)/4 ---> a(1)=(x+15)/16 The answer is E.
please explain me how x+3/4=4a(1)-3 , i uderstand your explanation just not this part
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Re: The sequence a(1), a(2), ..., a(n), ... is such that an=4a(n-1} - 3 [#permalink]
sujal998
ziko
The sequence a(1), a(2), …, a(n), … is such that a(n)=4a(n–1) –3 for all integers n>1. If a(3)=x, then a(1)=?

A. 4x–3
B. 16x–15
C. (x+3)/4
D. (x+3)/16
E. (x+15)/16

I love such questions!

Since we know a(3) we can find a(2), if we can find a(2) we can find a(1). a(3)=4a(2)-3 ---> x=4a(2)-3 ---> a(2)=(x+3)/4

a(2)=4a(1)-3 ---> (x+3)/4=4a(1)-3 ---> 4a(1)=(x+15)/4 ---> a(1)=(x+15)/16 The answer is E.

please explain me how x+3/4=4a(1)-3 , i uderstand your explanation just not this part

Substitute of $$a_2=\frac{x+3}{4}$$ into $$a_2=4a_1-3$$
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Re: The sequence a(1), a(2), ..., a(n), ... is such that an=4a(n-1} - 3 [#permalink]
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: The sequence a(1), a(2), ..., a(n), ... is such that an=4a(n-1} - 3 [#permalink]
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