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In a sequence of numbers in which each term after the first

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Eternal Intern
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In a sequence of numbers in which each term after the first [#permalink] New post 23 Jul 2003, 08:58
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A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

25% (00:00) correct 75% (00:10) wrong based on 5 sessions
In a sequence of numbers in which each term after the first term is 1 more than twice the preceding term, what is the fifth term?

(1) The first term is 1.
(2) The sixth term minus the fifth term is 32.

( 1) is definite right.

( 2) 2n( subscript 5) + 1- 2n( subscript 4) + 1
Should come out n5 - n4=32.
Now how can (2) be sufficient!

It is always use the stem: I can't remind you how important this is.
N5= 2N4 + 1
2N5- 2N4= 32-- ( 2)
2) Plug in
2( 2N4 + 1) - 2N4 =32
N4= 15
3) Plug back into variable
N5=31
Voila![/b]

Last edited by Curly05 on 23 Jul 2003, 10:11, edited 1 time in total.
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 [#permalink] New post 23 Jul 2003, 09:24
P = previous term
N = next term

for #(2), 6th term - 5th term = 32 is sufficient because,

P*2 + 1=N (holds for any 2 consecutive terms in this series) - Equation1

In this case, N - P = 32
=> N = P+32 -----euqation2

Solving equation 1 and equation 2, we get P = 31 which is the fifth term
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Answer [#permalink] New post 23 Jul 2003, 09:31
for choice (2), you must start from the beginning...

1st term= n
2nd = 2n+1
3rd =2(2n+1)+1 = 4n+5
4th = 2(4n+5)+1 = 8n+11
5th = 2(8n+11)+1 = 16n+23
6th = 2(16n+23)+1 = 32n+47

NOW... (32n+47) - (16n+23) = 32

solve that, n = 1, and that is suff (just like A)!!!!

Answer is D :-D
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Re: DS: Difficult Series Algebra [#permalink] New post 22 Mar 2011, 19:29
From(1)
T1 = 1, then T2 = 2*1 + 1 = 3 a series of Odd #, so fifth term can be found successively


From (2)

T6 - T5 = 32

T6 = 2T5 + 1

=> 2T5 + 1 - T5 = 32

=> T5 = 31

So Answer - D
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Re: DS: Difficult Series Algebra [#permalink] New post 22 Mar 2011, 19:42
Let xn be the nth term of the series...
xn=2x (n-1) +1
x2=2(x1) +1
x3=2(x2)+1=>4( x1)+3
x4=2(4(x1)+3)+1=>8(x1)+7
x5=2(8(x1)+7)+1=>16(x1)+15
x6=2(16(x1)+15)=32(x1)+30

Statement 1: x1=1 so x5= 16*1+15... Sufficient
Statement 2: 32(x1)+30-16(x1)-15=32..... Sufficient

D
Re: DS: Difficult Series Algebra   [#permalink] 22 Mar 2011, 19:42
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