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Quant Sequence Problem #131

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Quant Sequence Problem #131

by DCS80 » Sun Jan 27, 2013 11:50 am

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Can't follow this question for some reason. Maybe you guys can break it down and tailor it a bit better than the brief Quant explanation

131)

In the sequence X0, X1, X2...., Xn, each term from X1 to Xk is 3 greater than the previous term, and each term from Xk+1 to Xn is 3 less than the previous term, where n and k are positive integers and k < n. If X0 = Xn = 0 and if Xk = 15, what is the value of n?

A) 5
B) 6
C) 9
D) 10
E) 15

this question overwhelms me, and I understand sequences for the most part.


Thanks!!
Derek
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by GMATGuruNY » Sun Jan 27, 2013 12:45 pm
DCS80 wrote:Can't follow this question for some reason. Maybe you guys can break it down and tailor it a bit better than the brief Quant explanation

131)

In the sequence X0, X1, X2...., Xn, each term from X1 to Xk is 3 greater than the previous term, and each term from Xk+1 to Xn is 3 less than the previous term, where n and k are positive integers and k < n. If X0 = Xn = 0 and if Xk = 15, what is the value of n?

A) 5
B) 6
C) 9
D) 10
E) 15

this question overwhelms me, and I understand sequences for the most part.


Thanks!!
Derek
xâ‚€ = 0 and x(k) = 15.
From xâ‚€ to x(k), each term is 3 greater than the previous term.
Thus:
xâ‚€ = 0.
x� = 3.
xâ‚‚ = 6.
x₃ = 9.
xâ‚„ = 12.
xâ‚… = 15.
Since x(k) = 15, k=5.

x(k) = 15 and x(n) = 0.
From x(k+1) -- in other words, from x₆ -- to x(n), each term is 3 less than the previous term.
Thus:
x₆ = 12.
x₇ = 9.
x₈ = 6.
x₉ = 3.
x�₀ = 0.
Since x(n) = 0, n=10.

The correct answer is D.
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by ceilidh.erickson » Tue Jan 29, 2013 12:03 pm
The real difficulty with this problem is translating what they're really asking for. So try to break it down sentence by sentence:

each term from X1 to Xk is 3 greater than the previous term
Ok, this means that we're increasing by 3, until we get to a certain term "k."

each term from Xk+1 to Xn is 3 less than the previous term
So after the "kth" term, every term after that decreases by 3, until we get to the "nth" term

If X0 = Xn = 0 and if Xk = 15
Ok, if X0 and Xn are both 0, that means that we start and end at 0. Increase by 3 until we get to 15, then decrease by 3 til we're back to 0:
0, 3, 6, 9, 12, 15, 12, 9, 6, 3, 0

Here we have 11 terms. But as GMATGuru pointed out, the first term was actually X0, so the 11th term is X10. Answer: D.
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by Anaira Mitch » Tue Dec 11, 2018 3:51 pm
GMATGuruNY wrote:
DCS80 wrote:Can't follow this question for some reason. Maybe you guys can break it down and tailor it a bit better than the brief Quant explanation

131)

In the sequence X0, X1, X2...., Xn, each term from X1 to Xk is 3 greater than the previous term, and each term from Xk+1 to Xn is 3 less than the previous term, where n and k are positive integers and k < n. If X0 = Xn = 0 and if Xk = 15, what is the value of n?

A) 5
B) 6
C) 9
D) 10
E) 15

this question overwhelms me, and I understand sequences for the most part.


Thanks!!
Derek
xâ‚€ = 0 and x(k) = 15.
From xâ‚€ to x(k), each term is 3 greater than the previous term.
Thus:
xâ‚€ = 0.
x� = 3.
xâ‚‚ = 6.
x₃ = 9.
xâ‚„ = 12.
xâ‚… = 15.
Since x(k) = 15, k=5.

x(k) = 15 and x(n) = 0.
From x(k+1) -- in other words, from x₆ -- to x(n), each term is 3 less than the previous term.
Thus:
x₆ = 12.
x₇ = 9.
x₈ = 6.
x₉ = 3.
x�₀ = 0.
Since x(n) = 0, n=10.

The correct answer is D.
Thanks Mitch,

This is a very sweet solution.
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