Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

 It is currently 20 Jul 2019, 17:15 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # A certain sequence starts with term_1

Author Message
TAGS:

### Hide Tags

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

### Show Tags

3
Top Contributor
12 00:00

Difficulty:   85% (hard)

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

### HideShow timer Statistics 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

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

### Show Tags

4
Top Contributor
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

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
_________________
Senior Manager  G
Joined: 19 Apr 2016
Posts: 271
Location: India
GMAT 1: 570 Q48 V22 GMAT 2: 640 Q49 V28 GPA: 3.5
WE: Web Development (Computer Software)
A certain sequence starts with term_1  [#permalink]

### Show Tags

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  G
Joined: 24 Apr 2016
Posts: 327
Re: A certain sequence starts with term_1  [#permalink]

### Show Tags

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

Intern  B
Joined: 20 Feb 2017
Posts: 15
Location: India
GMAT 1: 560 Q47 V21 GPA: 3.05
WE: Other (Entertainment and Sports)
A certain sequence starts with term_1  [#permalink]

### Show Tags

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
Manager  S
Joined: 13 Dec 2013
Posts: 150
Location: United States (NY)
Schools: Cambridge"19 (A)
GMAT 1: 710 Q46 V41 GMAT 2: 720 Q48 V40 GPA: 4
WE: Consulting (Consulting)
Re: A certain sequence starts with term_1  [#permalink]

### Show Tags

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) = $$4K^2$$
Intern  B
Joined: 08 Dec 2016
Posts: 32
Location: Italy
Schools: IESE '21
Re: A certain sequence starts with term_1  [#permalink]

### Show Tags

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
Intern  B
Joined: 04 Jul 2018
Posts: 10
Re: A certain sequence starts with term_1  [#permalink]

### Show Tags

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
Display posts from previous: Sort by

# A certain sequence starts with term_1  