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

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Math Expert
Joined: 02 Sep 2009
Posts: 55271
In the sequence S, each term after the first is twice the previous  [#permalink]

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12 Jun 2015, 03:03
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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)

Kudos for a correct solution.

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In the sequence S, each term after the first is twice the previous  [#permalink]

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12 Jun 2015, 03:57
Bunuel wrote:
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)

Kudos for a correct solution.

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 $$S_{16}$$ = $$2^{13}*3*7$$ = $$2^{13}*21$$

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In the sequence S, each term after the first is twice the previous  [#permalink]

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13 Jun 2015, 03:55
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Bunuel wrote:
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)

Kudos for a correct solution.

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)

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Re: In the sequence S, each term after the first is twice the previous  [#permalink]

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13 Jun 2015, 05:15
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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= $$(3)(2^{13})$$+ $$(3)(2^{14})$$+$$(3)(2^{13})$$
=$$(3)(2^{13})[1+2+4]$$
= $$(21) (2^{13})$$

Option E
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Re: In the sequence S, each term after the first is twice the previous  [#permalink]

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13 Jun 2015, 09:50
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Bunuel wrote:
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)

Kudos for a correct solution.

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)
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Posts: 55271
Re: In the sequence S, each term after the first is twice the previous  [#permalink]

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15 Jun 2015, 02:59
Bunuel wrote:
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)

Kudos for a correct solution.

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.

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Re: In the sequence S, each term after the first is twice the previous  [#permalink]

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05 Mar 2017, 03:36
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)
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Re: In the sequence S, each term after the first is twice the previous  [#permalink]

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05 Apr 2019, 06:47
Top Contributor
Bunuel wrote:
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)

Kudos for a correct solution.

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)[1 + 2 + 2^2]
Simplify to get: (3)(2^13)[7]
Rewrite as: (21)(2^13)

Cheers,
Brent
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Re: In the sequence S, each term after the first is twice the previous   [#permalink] 05 Apr 2019, 06:47
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