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set 26 q 8

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by radhika1306 » Tue Sep 11, 2007 11:08 am
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: set 26 q 8

by wizardofwashington » Tue Sep 11, 2007 12:05 pm
radhika1306 wrote: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)
IMO, the answer is E.
The sequence can be rewritten as 1, 2^1, 2^2, 2^3......2^15, 2^16,2^17 (the last three listed here are the 16th, 17th and 18th terms, which are of our interest).

The sum of the 16,17 & 18th terms = 2^15+2^16+2^17
= 2^15 (1+2+2^2)
= 2^15 (7)
Hence our answer should be E.
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by kajcha » Tue Sep 11, 2007 12:09 pm
Agree E
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by radhika1306 » Thu Sep 13, 2007 9:29 am
correct thanks
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by camitava » Thu Sep 13, 2007 8:25 pm
This is a GP (Geometric progression) and in GP to find out the nth number, the formula is -

Code: Select all

Tn = ar^(n - 1)

- where a = first term and r = common ratio.
So from this, you can get the 16, 17 and 18th number and sum them up and it will give the answer - E.
Correct me If I am wrong


Regards,

Amitava
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