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by beater » Mon Sep 22, 2008 11:52 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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Source: — Problem Solving |

by jeenashiva » Mon Sep 22, 2008 12:03 pm
I will go with E.7(2 ^15).

What is the OA?
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by gmat579 » Mon Sep 22, 2008 12:47 pm
The sequence 1, 2, 4, 8, 16, 32, …, can be written as

2^0, 2^1, 2^2, 2^3,...

16th term is 2^15
17th term is 2^16
18th term is 2^17

Adding these - 2^15 + 2^16 + 2^17

= 1*2^15 + 2^1*2^15 + 2^2*2^15
= (1 + 2 + 4) 2^15
=7*2^15
Answer is E.
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by beater » Mon Sep 22, 2008 12:52 pm
Thanks guys! OA - E
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by pre-gmat » Mon Sep 22, 2008 1:05 pm
isnt this a geometric sequence?

what if we apply formula...

Sn = a1(1-r^n)/ 1-r
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by scoobydooby » Thu Sep 25, 2008 1:55 am
pre-gmat wrote:isnt this a geometric sequence?

what if we apply formula...

Sn = a1(1-r^n)/ 1-r

the above formula is for the sum of 1st n terms, we want only the sum of the 16th, 17th and the 18th term
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