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# PS: Geometric prog

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Director
Joined: 01 Apr 2008
Posts: 872
Name: Ronak Amin
Schools: IIM Lucknow (IPMX) - Class of 2014

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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)
Intern
Joined: 07 Feb 2009
Posts: 47

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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
Joined: 01 Apr 2008
Posts: 872
Name: Ronak Amin
Schools: IIM Lucknow (IPMX) - Class of 2014

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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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