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Another sequence problem

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Another sequence problem

by GmatTakerNo.1 » Mon May 03, 2010 10:47 am
Given a sequence 1, 2, 4, 8, 16, 32,..., each subsequent term after the first term is twice the previous term. What is the sum of the 16th 17th and 18th terms in the sequence?
a) 248
b) 3×217
c) 7×216
d) 3×216
e) 7×215

Answer is E. What is the fastest approach for this?
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by susantaiitk » Mon May 03, 2010 11:05 am
this should be easy.

Tn = 2^(n-1)

now easy to do the sum
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by GmatTakerNo.1 » Mon May 03, 2010 11:13 am
yeah ok, then I get 2^15+2^16+2^17

How can I symplify this to get 7*2^15?
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by iamseer » Mon May 03, 2010 11:30 am
something is wrong with your answer options.

sum of 16th, 17th, 18th term would be

2^15+2^16+ 2^17 = 2^15(1+2+4) = 2^15*7
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by GmatTakerNo.1 » Mon May 03, 2010 11:34 am
yeah thanks, I forgot the sign "^" between the 2 and the powers in the answer choices.
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by Stuart@KaplanGMAT » Mon May 03, 2010 3:52 pm
GmatTakerNo.1 wrote:yeah ok, then I get 2^15+2^16+2^17

How can I symplify this to get 7*2^15?
You simplify by factoring out 2^15 (the smallest power) from the 3 terms:

2^15 = 1*2^15
2^16 = 2^(1+15) = 2^1 * 2^15 = 2*2^15
2^17 = 2^(2+15) = 2^2 * 2^15 = 4*2^15

Adding them up:

1*2^15 + 2*2^15 + 4*2^15 = (1+2+4)*2^15 = 7*2^15
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by this_time_i_will » Mon May 03, 2010 4:53 pm
GmatTakerNo.1 wrote:Given a sequence 1, 2, 4, 8, 16, 32,..., each subsequent term after the first term is twice the previous term. What is the sum of the 16th 17th and 18th terms in the sequence?
a) 248
b) 3×217
c) 7×216
d) 3×216
e) 7×215

Answer is E. What is the fastest approach for this?
Just to generalize for any give GP:
Tn = a*r^(n-1); where a = first term, n = nth term and r is common ratio. for this problem common ratio = 2.
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by kstv » Tue May 04, 2010 5:51 pm
nth term in the series = 2^(n-1)
16th term = 2^(n-1) = 2^(16-1) = a
17th term = 2a , 18th term = 4a
Sum of 16th to 18th term = 7a = 7*2^15
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