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PS: Geometric prog

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Director
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Joined: 01 Apr 2008
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Schools: IIM Lucknow (IPMX) - Class of 2014
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PS: Geometric prog [#permalink] New post 22 Mar 2009, 22:09
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In the sequence 1, 2, 4, 8, 16, 32, …, each term after the first is twice the previous term.
What is the sum of the 16th, 17th, and 18th terms in the sequence?
A. 2^18
B. 3(2^17)
C. 7(2^16)
D. 3(2^16)
E. 7(2^15)
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Re: PS: Geometric prog [#permalink] New post 22 Mar 2009, 22:29
What is the OA?

I got E. 7(2^15)

Explanation:
As, T1 = 1
T2 = 2
T3 = 4

Tn = ar^(n-1), where a is the first term of the GP and r is the common ratio.
looking at the sequence, we can say that, a=1 and r=2.
Thus, Tn = 1* 2^(n-1)
or, Tn = 2^(n-1).

so, T16 = 2^(16-1) = 2^15.
similarly, T17 = 2^16
and T18 = 2^17

So,
Sum = T16+T17+T18 = 2^15+2^16+2^17
or, Sum = 2^15(1+2+2^2)
Or, Sum = 2^15(1+2+4)
Or, Sum = 7*2^15.
Director
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User avatar
Joined: 01 Apr 2008
Posts: 909
Schools: IIM Lucknow (IPMX) - Class of 2014
Followers: 15

Kudos [?]: 206 [0], given: 18

GMAT Tests User
Re: PS: Geometric prog [#permalink] New post 22 Mar 2009, 22:41
Perfect :)
I goofed up..applied Sum of n terms formula instead of Nth term formula :)
Thanks..
abhishekik wrote:
What is the OA?

I got E. 7(2^15)

Explanation:
As, T1 = 1
T2 = 2
T3 = 4

Tn = ar^(n-1), where a is the first term of the GP and r is the common ratio.
looking at the sequence, we can say that, a=1 and r=2.
Thus, Tn = 1* 2^(n-1)
or, Tn = 2^(n-1).

so, T16 = 2^(16-1) = 2^15.
similarly, T17 = 2^16
and T18 = 2^17

So,
Sum = T16+T17+T18 = 2^15+2^16+2^17
or, Sum = 2^15(1+2+2^2)
Or, Sum = 2^15(1+2+4)
Or, Sum = 7*2^15.
Re: PS: Geometric prog   [#permalink] 22 Mar 2009, 22:41
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