In the sequence 1, 1/2, 1/4, 1/8, 1/16, 1/32, ..., each term after the

Question

In the sequence 1, 1/2, 1/4, 1/8, 1/16, 1/32, ..., each term after the first is half the previous term. What is the sum of the 16th, 17th, and 18th terms?

A.  \frac{7}{2^{18}}

B.  \frac{7}{2^{17}}

C.  \frac{7}{2^{16}}

D.  \frac{7}{2^{15}}

E.  \frac{7}{2^{15}}

Gmatfinance · Answer

1/2^(n-1) if n =16 ,17 and 18

1/2^15 + 1/2^16 + 1/2^17 => 4/2^17 + 2/2^17 + 1/2^17 = 7/2^17 answer is B

ginevrafratto · Answer

can you please explain what have you done to obtain 7/2^17?

Harsh9676 · Answer

ginevrafratto

The sequence clearly is in Geometric progression (a, ar, ar2.....ar^n-1) in which a is the first element and r is the common ratio (Second element/First)

Given in the question r = 1/2 (you can derive it even if it is not mentioned using the above formula -- (1/2)/1 )

a = 1

We need the sum of 16th, 17th and 18th terms = a* r^ (16-1) + a * r ^ (17-1) + a * r ^ ( 18-1)

Factor out a and r^15 ,>> we get a* r ^15 * (1+r+r^2)

Now substitute a = 1 and r = 1/2 , >> 1 * (1/2)^15 * (1+1/2+(1/2)^2)

Solving this we get 7/2^17

Hope u understood

Gmatfinance · Answer

I first put the value of ns and in order to add 1/2^15 + 1/2^16 + 1/2^17, I multiplied 1/2^15 with 2^2/2^2 and 1/2^16 with 2/2 so I can rewrite them all as (value)/2^17 and add them.

In the sequence 1, 1/2, 1/4, 1/8, 1/16, 1/32, ..., each term after the

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GMAT Problem Solving (PS) Questions

In the sequence 1, 1/2, 1/4, 1/8, 1/16, 1/32, ..., each term after the

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