When I first looked at this problem I immediately looked to strategically guess, as any type of sequence question is out of my realm of capability. And after looking at the answer explanation, my guessing option was cemented.
In a geometric sequence each term is found by multiplying the previous term by a constant. If the first and second terms in a geometric sequence are 2x and 4x, what is the 600th term of the sequence?
a) (2^599)x
b) (2^600)x
c) 2^600x
d) (4^600)x
e) (4^599)x
Worth trying to solve?
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I believe that answer choice C should read as shown below:
Then:
a� = 2x = 2*2 = 4.
aâ‚‚ = 4x = 4*2 = 8.
Implication:
Since a₂ is twice as great as a�, each term in the sequence is twice the preceding term.
Thus:
a₃ = 2*8 = 16.
aâ‚„ = 2*16 = 32.
In this case, the sequence is composed of increasing powers of 2:
a� = 4 = 2².
a₂ = 8 = 2³.
a₃ = 16 = 2�.
a₄ = 32 = 2�.
Notice the pattern:
The exponent for the nth term is always equal to n+1.
Thus:
a₆₀₀ = 2��¹. This is our target.
Now plug x=2 into the answers to see which yields our target of 2��¹.
Only B works:
(2���)x = (2���)(2) = 2��¹.
The correct answer is B.
Let x=2.In a geometric sequence each term is found by multiplying the previous term by a constant. If the first and second terms in a geometric sequence are 2x and 4x, what is the 600th term of the sequence?
a) (2^599)x
b) (2^600)x
c) 2^(600x)
d) (4^600)x
e) (4^599)x
Then:
a� = 2x = 2*2 = 4.
aâ‚‚ = 4x = 4*2 = 8.
Implication:
Since a₂ is twice as great as a�, each term in the sequence is twice the preceding term.
Thus:
a₃ = 2*8 = 16.
aâ‚„ = 2*16 = 32.
In this case, the sequence is composed of increasing powers of 2:
a� = 4 = 2².
a₂ = 8 = 2³.
a₃ = 16 = 2�.
a₄ = 32 = 2�.
Notice the pattern:
The exponent for the nth term is always equal to n+1.
Thus:
a₆₀₀ = 2��¹. This is our target.
Now plug x=2 into the answers to see which yields our target of 2��¹.
Only B works:
(2���)x = (2���)(2) = 2��¹.
The correct answer is B.
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- Brent@GMATPrepNow
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The question says that each term is found by multiplying the previous term by a constantRastis wrote: In a geometric sequence each term is found by multiplying the previous term by a constant. If the first and second terms in a geometric sequence are 2x and 4x, what is the 600th term of the sequence?
a) (2^599)x
b) (2^600)x
c) 2^600x
d) (4^600)x
e) (4^599)x
NOTE the following:
term 1 = 2x
term 2 = 4x = (2)2x
So, the constant is 2
-----------------------
Now, let's list some terms and look for a pattern:
term 1 = 2x
term 2 = 2(2x) = 4x = (2²)x
term 3 = 2(4x) = 8x = (2³)x
term 4 = 2(8x) = 16x = (2�)x
term 5 = 2(16x) = 32x = (2�)x
.
.
.
term 600 = [spoiler](2^600)x[/spoiler]
Answer: B
Cheers,
Brent
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Hi Rastis,
Since each term = (previous term)(constant), we can compare the first two terms and see that the constant is 2, since 4x = (2x)(2). So, each term is simply a power of 2 greater than the one before:
First term = (2^1)x
Second term = (2^2)x
Third term = (2^3)x
...
600th term = (2^600)x
Let me know if you have any other questions about sequences -- they are not beyond you!
Since each term = (previous term)(constant), we can compare the first two terms and see that the constant is 2, since 4x = (2x)(2). So, each term is simply a power of 2 greater than the one before:
First term = (2^1)x
Second term = (2^2)x
Third term = (2^3)x
...
600th term = (2^600)x
Let me know if you have any other questions about sequences -- they are not beyond you!
Ready4
- Rastis
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I guess knowing to convert to exponents was also the big takeaway but nonetheless, I still would not know how to do this question if it appeared on the screen.
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The biggest take-away is to KEEP YOUR EYE ON THE ANSWER CHOICES.Rastis wrote:I guess knowing to convert to exponents was also the big takeaway but nonetheless, I still would not know how to do this question if it appeared on the screen.
Since the answer choices are phrased in terms of exponents, we should phrase the sequence in terms of exponents.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
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