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Question from Gmat Prep- Help please! 6

Expert replies
by bobdylan » Sat Apr 14, 2012 6:26 am
In the arithmetic sequence t1,t2, t3...tn, t1=23 and tn=tn-1 -3 for each n>1. What's the value of n when tn= -4.
a)-1
b)7
c)10
d)14
e)20
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Source: — Problem Solving |

by GMATGuruNY » Sat Apr 14, 2012 6:44 am
bobdylan wrote:In the arithmetic sequence t1,t2, t3...tn, t1=23 and tn= t(n-1)-3 for each n>1. What's the value of n when tn= -4.
a)-1
b)7
c)10
d)14
e)20
Since tn= t(n-1)-3, the distance between one value and the next is 3.
Thus, the terms in the sequence are EVENLY SPACED, with an INTERVAL of 3 between successive values.
To count evenly spaced integers, use the following formula:

Number of integers = (biggest-smallest)/interval + 1

In the problem at hand:
Number of integers = (23-(-4))/3 + 1 = 10.
Since there are 10 terms, t(n) = -4 when n=10.

The correct answer is C.
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by Shalabh's Quants » Mon Apr 16, 2012 6:01 am
bobdylan wrote:In the arithmetic sequence t1,t2, t3...tn, t1=23 and tn=tn-1 -3 for each n>1. What's the value of n when tn= -4.
a)-1
b)7
c)10
d)14
e)20
For an AP... tn = t1 + (n-1).d; where tn is last term = -4, t1= I term = 23, n = no. of terms, d = common difference = -3.

So, -4 = 23 +(n-1)*-3 => n = 10.
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by Anurag@Gurome » Mon Apr 16, 2012 6:49 am
bobdylan wrote:In the arithmetic sequence t1,t2, t3...tn, t1=23 and tn=tn-1 -3 for each n>1. What's the value of n when tn= -4.
a)-1
b)7
c)10
d)14
e)20

t(1) = 23 and t(n) = t(n - 1) - 3 for each n > 1

Hence, the series is an arithmetic progression with first term, a = 23 and common difference, d = -3.

Hence, t(n) = a + (n - 1)d = 23 + (n - 1)*(-3) = 23 - 3n + 3 = 26 - 3n

Given that, t(n) = -4
(26 - 3n) = -4
3n = 30
n = 10

The correct answer is C.
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by ronnie1985 » Mon Apr 16, 2012 7:03 am
(C) QED
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by aneesh.kg » Mon Apr 16, 2012 9:10 am
Since this is an AP, the difference between tn and tn-1 must be the common difference (d).
Given that tn - tn-1 = -3, therefore d = -3.

(Note that it is an decreasing AP, so it should be no surprise that the given nth term is smaller than the given first term.)

A term at the nth position of an AP, tn = t1 + (n-1).d,
where
t1 = 23 (given)
tn = -4
d = -3

Upon plugging these values in the above expression for tn, we get n = 10.

[C] is the answer.
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