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# A certain sequence starts with term_1

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CEO
Joined: 12 Sep 2015
Posts: 3853
A certain sequence starts with term_1  [#permalink]

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06 Apr 2017, 08:07
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Difficulty:

85% (hard)

Question Stats:

58% (02:54) correct 42% (03:05) wrong based on 162 sessions

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A certain sequence starts with term_1

For any term in the sequence, term_n = 16^(2n - 1)

If the PRODUCT of the first k terms of the sequence is 2^1600, what is the value of k?

A) 5
B) 10
C) 20
D) 40
E) 80

* Kudos for all correct solutions

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Joined: 12 Sep 2015
Posts: 3853
Re: A certain sequence starts with term_1  [#permalink]

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07 Apr 2017, 07:34
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Top Contributor
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GMATPrepNow wrote:
A certain sequence starts with term_1

For any term in the sequence, term_n = 16^(2n - 1)

If the PRODUCT of the first k terms of the sequence is 2^1600, what is the value of k?

A) 5
B) 10
C) 20
D) 40
E) 80

First notice that 2n - 1, will be ODD for all integer values of n. For example:
If n = 1, then 2n - 1 = 2(1) - 1 = 1
If n = 2, then 2n - 1 = 2(2) - 1 = 3
If n = 3, then 2n - 1 = 2(3) - 1 = 5
If n = 4, then 2n - 1 = 2(4) - 1 = 7
.
.
.
etc.

Now notice what happens when we add consecutive ODD numbers (starting with 1)
The first 1 ODD number: 1 = 1 (and 1 = 1²)
The first 2 ODD numbers: 1 + 3 = 4 (and 4 = 2²)
The first 3 ODD numbers: 1 + 3 + 5 = 9 (and 9 = 3²)
The first 4 ODD numbers: 1 + 3 + 5 + 7 = 16 (and 16 = 4²)
The first 5 ODD numbers: 1 + 3 + 5 + 7 + 9 = 25 (and 25 = 5²)
.
.
.
In general, the sum of the first k ODD numbers = k²

Now onto the question!!!

term_n = 16^(2n - 1)
term_1 = 16^(2(1) - 1) = 16^1
term_2 = 16^(2(2) - 1) = 16^3
term_3 = 16^(3(3) - 1) = 16^5
term_4 = 16^(2(4) - 1) = 16^7
etc

So, the PRODUCT of the first k terms = (16^1)(16^3)(16^5)(16^7)(16^9). . . (16^??)
When we multiply powers with the same base, we ADD the exponents.
So, the PRODUCT of the first k terms = 16^(1 + 3 + 5 + 7 + . . . ??)

Notice that the exponent here is equal to the SUM of the first k ODD numbers.
Well, we already know that the sum of the first k ODD numbers = k²
So, the PRODUCT of the first k terms = 16^()

We're told that the PRODUCT of the first k terms is 2^1600
So, we can write: 16^() = 2^1600

We need the same base, so let's rewrite 16 as 2^4
We get: (2^4)^() = 2^1600
Apply power of a power law: 2^(4k²) = 2^1600
This means that 4k² = 1600
Divide both sides by 4 to get: k² = 400
Solve: k = 20 or -20
Since -20 makes no sense, we know that k = 20

In other words, the PRODUCT of the first 20 terms of the sequence is 2^1600,

Cheers,
Brent
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A certain sequence starts with term_1  [#permalink]

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Updated on: 06 Apr 2017, 08:55
7
2
GMATPrepNow wrote:
A certain sequence starts with term_1

For any term in the sequence, term_n = 16^(2n - 1)

If the PRODUCT of the first k terms of the sequence is 2^1600, what is the value of k?

A) 5
B) 10
C) 20
D) 40
E) 80

* Kudos for all correct solutions

term_n = 16^(2n - 1) = $$2^{8n-4}$$

term_1 = $$2^{8-4} = 2^4$$

term_2 = $$2^{16-4} = 2^{12}$$

term_3 = $$2^{24-4} = 2^{20}$$
.
.
.
term_k = $$2^{8k-4}$$

product of the first k terms of the sequence is 2^1600

term_1 * term_2 * term_3 ......* term_k =$$2^{1600}$$

$$2^4*2^{12}*2^{20}.....*2^{8k-4}=2^{1600}$$

$$2^{4+12+20+....+8k-4}=2^{1600}$$

4+12+20+...8k-4 = 1600

sum of the above AP = (k/2)(4+8k-4) = $$4K^2$$

$$4K^2=1600$$

$$K^2=400$$

k=20

Hence option C is correct
Hit Kudos if you liked it

Originally posted by 0akshay0 on 06 Apr 2017, 08:54.
Last edited by 0akshay0 on 06 Apr 2017, 08:55, edited 1 time in total.
##### General Discussion
Senior Manager
Joined: 24 Apr 2016
Posts: 327
Re: A certain sequence starts with term_1  [#permalink]

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06 Apr 2017, 08:53
1
1
term_1 = 16 ^ (2-1) = 16 ^ 1
term_2 = 16 ^ (4-1) = 16 ^ 3
term_3 = 16 ^ (6-1) = 16 ^ 5

So we see a pattern forming.

question says that (16 ^ 1) * (16 ^ 3) * (16 ^ 5) *.... [16 ^ (2k-1)] = 2^1600 = 16 ^ 400

hence the powers of 16 on the left side add up to 400

1+3+5 +....2k-1 = 400

Using the formula :: n/2 [2+(n-1)2] = 400

n/2[2+2n-2] = 400 ==> n/2[2n] = 400 ==> n^2 = 20^2 ==> n=20

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A certain sequence starts with term_1  [#permalink]

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06 Apr 2017, 08:55
3
Since,

nth term = term_n = 16^(2n - 1), the series will be 16,16^3,16^5,16^7,.....

Product of 2 terms = 16^4
Product of 3 terms = 16^9
Product of 4 terms = 16^16
Product of n terms = 16^n^2

2^1600 = 16^400
ie; n^2 = 400, so n or here k = 20
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Re: A certain sequence starts with term_1  [#permalink]

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12 Apr 2017, 17:33
1
0akshay0 wrote:
GMATPrepNow wrote:
A certain sequence starts with term_1

For any term in the sequence, term_n = 16^(2n - 1)

If the PRODUCT of the first k terms of the sequence is 2^1600, what is the value of k?

A) 5
B) 10
C) 20
D) 40
E) 80

* Kudos for all correct solutions

term_n = 16^(2n - 1) = $$2^{8n-4}$$

term_1 = $$2^{8-4} = 2^4$$

term_2 = $$2^{16-4} = 2^{12}$$

term_3 = $$2^{24-4} = 2^{20}$$
.
.
.
term_k = $$2^{8k-4}$$

product of the first k terms of the sequence is 2^1600

term_1 * term_2 * term_3 ......* term_k =$$2^{1600}$$

$$2^4*2^{12}*2^{20}.....*2^{8k-4}=2^{1600}$$

$$2^{4+12+20+....+8k-4}=2^{1600}$$

4+12+20+...8k-4 = 1600

sum of the above AP = (k/2)(4+8k-4) = $$4K^2$$

$$4K^2=1600$$

$$K^2=400$$

k=20

Hence option C is correct
Hit Kudos if you liked it

Thanks for the post. Can you go over this part?

sum of the above AP = (k/2)(4+8k-4) = $$4K^2$$
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Re: A certain sequence starts with term_1  [#permalink]

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10 May 2017, 07:08
1
Cez005 wrote:
0akshay0 wrote:
GMATPrepNow wrote:
A certain sequence starts with term_1

For any term in the sequence, term_n = 16^(2n - 1)

If the PRODUCT of the first k terms of the sequence is 2^1600, what is the value of k?

A) 5
B) 10
C) 20
D) 40
E) 80

* Kudos for all correct solutions

term_n = 16^(2n - 1) = $$2^{8n-4}$$

term_1 = $$2^{8-4} = 2^4$$

term_2 = $$2^{16-4} = 2^{12}$$

term_3 = $$2^{24-4} = 2^{20}$$
.
.
.
term_k = $$2^{8k-4}$$

product of the first k terms of the sequence is 2^1600

term_1 * term_2 * term_3 ......* term_k =$$2^{1600}$$

$$2^4*2^{12}*2^{20}.....*2^{8k-4}=2^{1600}$$

$$2^{4+12+20+....+8k-4}=2^{1600}$$

4+12+20+...8k-4 = 1600

sum of the above AP = (k/2)(4+8k-4) = $$4K^2$$

$$4K^2=1600$$

$$K^2=400$$

k=20

Hence option C is correct
Hit Kudos if you liked it

Thanks for the post. Can you go over this part?

sum of the above AP = (k/2)(4+8k-4) = $$4K^2$$

-------------------------------------------------------------------------
Formula: #terms(Val_max+Val_min)/2 = sum of equidistant numbers
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Joined: 04 Jul 2018
Posts: 10
Re: A certain sequence starts with term_1  [#permalink]

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21 Aug 2018, 05:46
quantumliner wrote:
term_1 = 16 ^ (2-1) = 16 ^ 1
term_2 = 16 ^ (4-1) = 16 ^ 3
term_3 = 16 ^ (6-1) = 16 ^ 5

So we see a pattern forming.

question says that (16 ^ 1) * (16 ^ 3) * (16 ^ 5) *.... [16 ^ (2k-1)] = 2^1600 = 16 ^ 400

hence the powers of 16 on the left side add up to 400

1+3+5 +....2k-1 = 400

Using the formula :: n/2 [2+(n-1)2] = 400

n/2[2+2n-2] = 400 ==> n/2[2n] = 400 ==> n^2 = 20^2 ==> n=20

How do I get the formula : n/2 [2+(n-1)2] = 400 ???
Is there any previous post that explains this formula?
Re: A certain sequence starts with term_1   [#permalink] 21 Aug 2018, 05:46
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