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Worth trying to solve?

This topic has 4 expert replies and 1 member reply
Rastis Master | Next Rank: 500 Posts
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Worth trying to solve?

Post Tue Mar 24, 2015 7:32 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    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

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    Post Tue Mar 24, 2015 8:02 am
    I believe that answer choice C should read as shown below:

    Quote:
    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
    Let x=2.
    Then:
    a₁ = 2x = 2*2 = 4.
    a₂ = 4x = 4*2 = 8.

    Implication:
    Since a₂ is twice as great as a₁, each term in the sequence is twice the preceding term.
    Thus:
    a₃ = 2*8 = 16.
    a₄ = 2*16 = 32.

    In this case, the sequence is composed of increasing powers of 2:
    a₁ = 4 = 2².
    a₂ = 8 = 2³.
    a₃ = 16 = 2⁴.
    a₄ = 32 = 2⁵.

    Notice the pattern:
    The exponent for the nth term is always equal to n+1.
    Thus:
    a₆₀₀ = 2⁶⁰¹. This is our target.

    Now plug x=2 into the answers to see which yields our target of 2⁶⁰¹.
    Only B works:
    (2⁶⁰⁰)x = (2⁶⁰⁰)(2) = 2⁶⁰¹.

    The correct answer is B.

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    Post Tue Mar 24, 2015 8:09 am
    Rastis 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
    The question says that each term is found by multiplying the previous term by a constant
    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 = (2⁴)x
    term 5 = 2(16x) = 32x = (2⁵)x
    .
    .
    .
    term 600 = (2^600)x

    Answer: B

    Cheers,
    Brent

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    GMAT/MBA Expert

    Spencer@Prep4GMAT Junior | Next Rank: 30 Posts
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    Post Tue Mar 24, 2015 8:09 am
    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! Smile

    _________________
    Ready4

    Rastis Master | Next Rank: 500 Posts
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    Posted:
    183 messages
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    2 members
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    Target GMAT Score:
    700
    GMAT Score:
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    Post Tue Mar 24, 2015 8:18 am
    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.

    Post Tue Mar 24, 2015 8:22 am
    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.
    The biggest take-away is to KEEP YOUR EYE ON THE ANSWER CHOICES.
    Since the answer choices are phrased in terms of exponents, we should phrase the sequence in terms of exponents.

    _________________
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    If you find one of my posts helpful, please take a moment to click on the "Thank" icon.
    Available for tutoring in NYC and long-distance.
    For more information, please email me at GMATGuruNY@gmail.com.

    Thanked by: Rastis
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