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
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Difficulty level: 700+
TRICKY sequence question
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Brent original!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+
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
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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.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+
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
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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
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First notice that 2n - 1, will be ODD for all integer values of n. For example: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
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