In the sequence S, each term after the first is twice the previous

Question

In the sequence S, each term after the first is twice the previous term. If the first term of sequence S is 3, what is the sum of the 14th, 15th, and 16th terms in sequence S?

(A) 3(2^16)
(B) 9(2^15)
(C) 21(2^14)
(D) 9(2^14)
(E) 21(2^13)

GMATinsight · Accepted Answer

Given : In the sequence S, each term after the first is twice the previous term  
i.e.
 S_1 = 3 
 S_2 = 2*3
S_3 = 2^2*3
S_4 = 2^3*3 
...
...
 S_{14} = 2^{13}*3 
 S_{15} = 2^{14}*3 
 S_{16} = 2^{15}*3

Sum of  S_{14} ,  S_{15}  and  S_{16}  =  2^{13}*3 + 2^{14}*3 + 2^{15}*3

i.e. Sum of  S_{14} ,  S_{15}  and  S_{16}  =  2^{13}*3 ( 1+ 2 + 2^2)

i.e. Sum of  S_{14} ,  S_{15}  and  S_{16}  =  2^{13}*3*7  =  2^{13}*21

Answer: Option  E

ritigl · Answer

This is a geometric progression with common ratio (r) = 2
First term (a) = 3

nth Term in Geometric Progression =  ar^{(n-1)}  
Hence, 
14th term= (3)(2^{13}) 
15th term=  (3)(2^{14}) 
16th term=  (3)(2^{15})

Sum of the above three terms=  (3)(2^{13}) +  (3)(2^{14}) + (3)(2^{13}) 
= (3)(2^{13})  
=  (21) (2^{13})

Option E

ShristiK · Answer

The terms in the sequence can be shown as
a(n) = 2* a(n-1)
So, the sequence will look like: 3, 2*3, (2^2) *3,...

And the nth term will be given as 2^(n-1) *3

Therefore, a(14) = (2^13)*3, a(15) = (2^14) *3, and a(16) = (2^15) *3

So, a(14) + a(15) + a(16) = (2^13) *3 + (2^14) *3 + (2^15) *3
 = 3* (2^13) *(1+2+4) = 3* (2^13) *7
 = 21 * (2^13)

Answer : E

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KS15 · Answer

First Term=3
Second Term=3*2
Third Term=3*(2^2)
Similarly,
Sum of 14th, 15th and 16th terms= 3*(2^13)+3*(2^14)+3*(2^15)=3*(2^13) {1+2+2^2}=3*7*(2^13)=21*(2^13)
Answer E

Bunuel · Answer

MANHATTAN GMAT OFFICIAL SOLUTION:

The sequence S is 3, 6, 12, 24, 48, and so on. We could write out the first 16 terms and add the 14th, 15th, and 16th together, but such an approach would be prone to error and time consuming. Additionally, we don't need to calculate the sum explicitly: the answer choices all have some power of 2 as a factor, providing a hint at the best solution method.

Let's write the sequence in terms of the powers of 2:
S: 3, 6, 12, 24, 48, and so on.
S: 3(2^0), 3(2^1), 3(2^2), 3(2^3), 3(2^4), and so on.
So Sn, the nth term of S, equals 3*2^(n - 1).

Thus, the sum of the 14th, 15th, and 16th terms equals 3(2^13) + 3(2^14) + 3(2^15).

All of the terms share the common factors 3 and 2^13, so factor those terms out:

3(2^13)(1 + 2^1 + 2^2) = 21*2^13.

The correct answer is E

anairamitch1804 · Answer

The sequence S is 3, 6, 12, 24, 48, and so on. We could write out the first 16 terms and add the 14th, 15th, and 16th together, but such an approach would be prone to error and time consuming. Additionally, we don't need to calculate the sum explicitly: the answer choices all have some power of 2 as a factor, providing a hint at the best solution method.
Let's write the sequence in terms of the powers of 2:
S: 3, 6, 12, 24, 48, and so on.
S: 3(2^0), 3(2^1), 3(2^2), 3(2^3), 3(2^4), and so on.
So Sn, the nth term of S, equals 3(2^n – 1).
Thus, the sum of the 14th, 15th, and 16th terms equals 3(2^13) + 3(2^14) + 3(2^15).
All of the terms share the common factors 3 and 213, so factor those terms out:
3(2^13)(1 + 2^1 + 2^2)
3(2^13)(1 + 2 + 4)
3(2^13)(7)
21(2^13)
The correct answer is E.

BrentGMATPrepNow · Answer

term1 = 3
term2 = (3)(2) = (3)(2¹) 
term3 = (3)(2)(2) = (3)(2²) 
term4 = (3)(2)(2)(2) = (3)(2³)
term5 = (3)(2)(2)(2)(2) = (3)(2⁴)
.
.
term14 = (3)(2^13)
term15 = (3)(2^14)
term16 = (3)(2^15)

What is the sum of the 14th, 15th, and 16th terms in sequence S?  
Sum = (3)(2^13) + (3)(2^14) + (3)(2^15)
Factor out 3(2^13) to get: (3)(2^13) 
Simplify to get: (3)(2^13) 
Rewrite as: (21)(2^13)

Answer: E

Cheers, 
Brent

ScottTargetTestPrep · Answer

Let's write the first few terms of this sequence:

\bullet  S1 = 3
 \bullet  S2 = 2 * 3
 \bullet  S3 = 2 * 2 * 3 = 2^2 * 3
 \bullet  S4 = 2 * 2^2 * 3 = 2^3 * 3

Notice that the exponent of 2 in each term is one less than the index of that term. Thus, the exponent of 2 in the nth term will be n - 1, i.e. Sn = 2^(n - 1) * 3.

Using this formula, we can write S14 = 2^13 * 3, S15 = 2^14 * 3, and S16 = 2^15 * 3. Let's add these terms:

\Rightarrow  S14 + S15 + S16

\Rightarrow  2^13 * 3 + 2^14 * 3 + 2^15 * 3

\Rightarrow  3(2^13 + 2^14 + 2^15)

\Rightarrow  3 * 2^13 * (1 + 2 + 4)

\Rightarrow  3 * 2^13 * 7

\Rightarrow  21 * 2^13

Answer: E

In the sequence S, each term after the first is twice the previous

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