TRICKY sequence question

This topic has expert replies

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

TRICKY sequence question

by Brent@GMATPrepNow » Thu Apr 06, 2017 7:09 am
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

Answer: C

Source: www.gmatprepnow.com
Difficulty level: 700+
Brent Hanneson - Creator of GMATPrepNow.com
Image

User avatar
Legendary Member
Posts: 2663
Joined: Wed Jan 14, 2015 8:25 am
Location: Boston, MA
Thanked: 1153 times
Followed by:128 members
GMAT Score:770

by DavidG@VeritasPrep » Thu Apr 06, 2017 8:11 am
Brent@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

Answer: C

Source: www.gmatprepnow.com
Difficulty level: 700+
Brent original!

Let's see if we can suss out a pattern here. (We'll do this by plugging values of n into 16^(2n -1)

The first term = 16 = 2^4
The second term = 16^3 = 2^12
The third term = 16^5 = 2^20
The fourth term = 16^7 = 2^28

So we're starting with 2^4, and with each iteration, we're increasing the exponent by 8.

So the first term is 2^4. For the second term we're adding a one eight to the exponent. For the third term, we're adding two eights to the exponent. Thus, for the nth term, we'd add (n-1) 8's to the exponent.

We want a total sum of 1600 in the exponent. So, essentially, what we want is 4 + 12 + 20... = 1600

Our low term is 4.
If our high term is n, then we'll add 8 a total of n-1 times, giving us 4 + 8(n-1).
And our total number of terms = n.

Because this is an evenly spaced set, (the gap between each term is 8) we know that the average of the set is (High + Low)/2. In this case:
[4 + 8(n-1) + 4]/2 = [4 + 4 + 8n - 8]/2 = 8n/2 = 4n. So 4n is the average.

We know that average * number of terms = sum.
4n is the average, n is the number of terms, and 1600 is the sum.
Thus we have: 4n *n = 1600. 4n^2 = 1600 ---> n^2 = 400 ---> n = 20. The answer is C
Veritas Prep | GMAT Instructor

Veritas Prep Reviews
Save $100 off any live Veritas Prep GMAT Course

User avatar
Legendary Member
Posts: 2663
Joined: Wed Jan 14, 2015 8:25 am
Location: Boston, MA
Thanked: 1153 times
Followed by:128 members
GMAT Score:770

by DavidG@VeritasPrep » Thu Apr 06, 2017 8:15 am
Brent@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

Answer: C

Source: www.gmatprepnow.com
Difficulty level: 700+
We could also back-solve. Let's test B. Using the above analysis, if n = 10, and the first term in the series is 4, we'd add 8 a total nine times, to get 4 + 8*9 = 76. Our average would be (4 + 76)/2 = 40.
If there are 10 terms, our sum would be 40*10 = 400. Too small. We want 1600.

Test D. If n = 40, and the first term in the series is 4, we'd add 8 a total thirty-nine times, to get 4 + 8*39 = 316. Our average would be (4 + 316)/2 = 160.
If there are 40 terms, our sum would be 160*40 = 6400 Too big.

If B is too small and D is too big, the answer has to be C
Veritas Prep | GMAT Instructor

Veritas Prep Reviews
Save $100 off any live Veritas Prep GMAT Course

User avatar
Legendary Member
Posts: 2663
Joined: Wed Jan 14, 2015 8:25 am
Location: Boston, MA
Thanked: 1153 times
Followed by:128 members
GMAT Score:770

by DavidG@VeritasPrep » Thu Apr 06, 2017 11:20 am
For another good question involving evenly spaced sets (and the axiom that the mean = median in such sets) see here: https://www.beatthegmat.com/gmat-prep-ar ... 10553.html
Veritas Prep | GMAT Instructor

Veritas Prep Reviews
Save $100 off any live Veritas Prep GMAT Course

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Fri Apr 07, 2017 6:33 am
Brent@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^(k²)

We're told that the PRODUCT of the first k terms is 2^1600
So, we can write: 16^(k²) = 2^1600

We need the same base, so let's rewrite 16 as 2^4
We get: (2^4)^(k²) = 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,

Answer: C

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image