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OG PS #171 - Easy Sequence Question!

This topic has 1 expert reply and 3 member replies
kalifalk Newbie | Next Rank: 10 Posts
Joined
25 Nov 2009
Posted:
4 messages

OG PS #171 - Easy Sequence Question!

Post Thu Dec 15, 2011 8:01 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    Hello all,

    I've never posted before, but I have a simple question about OG PS question #171. Question is:

    "What is the difference between the sixth and the fifth terms of the sequence 2, 4, 7.... whose nth term is n+2^n-1."

    According to the answer, you simply plug 6 and 5 into the term formula, but my question is why did they list seemingly random numbers first... I had assumed that those were the 1st, 2nd, and 3rd terms, I tried to plug them in as recursive numbers.

    If someone could explain that would be fantastic Smile

    Thanks guys! Happy studying and/or happy tutoring!

    Kali

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    LalaB Master | Next Rank: 500 Posts
    Joined
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    Posted:
    425 messages
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    GMAT Score:
    690
    Post Thu Dec 15, 2011 8:14 am
    kalifalk wrote:
    Hello all,

    I've never posted before, but I have a simple question about OG PS question #171. Question is:

    "What is the difference between the sixth and the fifth terms of the sequence 2, 4, 7.... whose nth term is n+2^n-1."

    According to the answer, you simply plug 6 and 5 into the term formula, but my question is why did they list seemingly random numbers first... I had assumed that those were the 1st, 2nd, and 3rd terms, I tried to plug them in as recursive numbers.

    If someone could explain that would be fantastic Smile

    Thanks guys! Happy studying and/or happy tutoring!

    Kali
    numbers are not random. plug in them to the formula n+2^n-1

    so, ur 1st term will be equal to 1+2^1-1=1+1=2 (as it is mentioned in ur sequence)
    ur 2nd term is 2+2^2-12+2=4
    ur 3d term is 3+2^3-1=3+4=7

    the same sequence (2, 4, 7....) is listed in the q.stem

    hope it helps

    ashishpks Newbie | Next Rank: 10 Posts Default Avatar
    Joined
    06 Feb 2012
    Posted:
    3 messages
    Post Sat Apr 28, 2012 5:26 am
    Thanks Buddy

    greatsaint Junior | Next Rank: 30 Posts Default Avatar
    Joined
    18 Mar 2009
    Posted:
    16 messages
    Post Sat Apr 28, 2012 5:37 am
    kalifalk wrote:
    Hello all,

    I've never posted before, but I have a simple question about OG PS question #171. Question is:

    "What is the difference between the sixth and the fifth terms of the sequence 2, 4, 7.... whose nth term is n+2^n-1."

    According to the answer, you simply plug 6 and 5 into the term formula, but my question is why did they list seemingly random numbers first... I had assumed that those were the 1st, 2nd, and 3rd terms, I tried to plug them in as recursive numbers.

    If someone could explain that would be fantastic Smile

    Thanks guys! Happy studying and/or happy tutoring!

    Kali
    The first term in the sequnce is n=0. So for the 5th term, n=4 and 6th term , n=5.
    Thus your answer comes to (4+2)^(4-1) which 6^3 and the 6th term should be (5+2)^(5-1) which is 7^4.

    Here, as you notice value of n is 1 less than the term of the sequnce (Eg: Sixth term is n=5). I feel this is the trick that they are testing.

    Post Fri Jul 10, 2015 4:36 am
    kalifalk wrote:
    Hello all,

    I've never posted before, but I have a simple question about OG PS question #171. Question is:

    "What is the difference between the sixth and the fifth terms of the sequence 2, 4, 7.... whose nth term is n+2^n-1."

    According to the answer, you simply plug 6 and 5 into the term formula, but my question is why did they list seemingly random numbers first... I had assumed that those were the 1st, 2nd, and 3rd terms, I tried to plug them in as recursive numbers.

    If someone could explain that would be fantastic Smile

    Thanks guys! Happy studying and/or happy tutoring!

    Kali
    What is the difference between the sixth and the fifth terms of the sequence 2, 4, 7, ... whose nth term is n + 2^(n - 1)?

    (A) 2
    (B) 3
    (C) 6
    (D) 16
    (E) 17

    Solution:

    To find the sixth term, let's first substitute 6 for n.

    6 + 2^(6-1)

    6 + 2^5

    6 + 32 = 38

    To find the fifth term, we substitute 5 for n.

    5 + 2^(5-1)

    5 + 2^4

    5 + 16 = 21

    The difference between the 6th and 5th terms of the sequence is 38 - 21 = 17.

    Answer: E

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