In the arithmetic sequence t1, t2, t3, ..., tn, t1=23 and tn= tn-1 - 3 for each n > 1. What is the value of n when tn = -4?
A. -1
B. 7
C. 10
D. 14
E. 20
I know the answer is 10 and have seen some solutions around. I'm actually looking for an easy and faster way to solve that problem.
Thanks.
in the arithmetic sequence
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t(n) = t(n) - 3.didieravoaka wrote:In the arithmetic sequence t1, t2, t3, ..., tn, t1=23 and tn= tn-1 - 3 for each n > 1. What is the value of n when tn = -4?
A. -1
B. 7
C. 10
D. 14
E. 20
According to the formula above, each term is 3 less than the preceding term.
The result is an EVENLY SPACED SET.
The value of n when t(n) = -4 is equal to the NUMBER OF TERMS when the biggest term is 23 and the smallest term is -4.
To count the number of integers in an evenly spaced set, use the following formula:
(biggest - smallest)/interval + 1.
The interval is the distance between successive terms.
Here, the interval is 3, since each term is 3 less than the preceding term.
Since the biggest term = 23 and the smallest term = -4, we get:
Number of terms = (biggest - smallest)/interval + 1 = [23 - (-4)]/3 + 1 = 10.
The correct answer is C.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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GMATGuruNY wrote:t(n) = t(n) - 3.didieravoaka wrote:In the arithmetic sequence t1, t2, t3, ..., tn, t1=23 and tn= tn-1 - 3 for each n > 1. What is the value of n when tn = -4?
A. -1
B. 7
C. 10
D. 14
E. 20
According to the formula above, each term is 3 less than the preceding term.
The result is an EVENLY SPACED SET.
The value of n when t(n) = -4 is equal to the NUMBER OF TERMS when the biggest term is 23 and the smallest term is -4.
To count the number of integers in an evenly spaced set, use the following formula:
(biggest - smallest)/interval + 1.
The interval is the distance between successive terms.
Here, the interval is 3, since each term is 3 less than the preceding term.
Since the biggest term = 23 and the smallest term = -4, we get:
Number of terms = (biggest - smallest)/interval + 1 = [23 - (-4)]/3 + 1 = 10.
The correct answer is C.
Really really really easy and helpful.
Thanks GuruNY
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Mitch's approach is nice and fast.didieravoaka wrote:In the arithmetic sequence t1, t2, t3, ..., tn, t1=23 and tn= tn-1 - 3 for each n > 1. What is the value of n when tn = -4?
A. -1
B. 7
C. 10
D. 14
E. 20
However, if you didn't spot that approach, there's another approach that might be just as fast:
We're told that each term is 3 less than the preceding term.
So, we have:
term1 = 23
term2 = 20
term3 = 17
term4 = 14
term5 = 11
term6 = 8
term7 = 5
term8 = 2
term9 = -1
term10 = -4
Answer: C
Yes, this solution is not very "mathematical" but the SOLE GOAL with GMAT questions is to answer the question AS ACCURATELY AND AS FAST AS POSSIBLE.
Cheers,
Brent
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Thanks Brent,Brent@GMATPrepNow wrote:Mitch's approach is nice and fast.didieravoaka wrote:In the arithmetic sequence t1, t2, t3, ..., tn, t1=23 and tn= tn-1 - 3 for each n > 1. What is the value of n when tn = -4?
A. -1
B. 7
C. 10
D. 14
E. 20
However, if you didn't spot that approach, there's another approach that might be just as fast:
We're told that each term is 3 less than the preceding term.
So, we have:
term1 = 23
term2 = 20
term3 = 17
term4 = 14
term5 = 11
term6 = 8
term7 = 5
term8 = 2
term9 = -1
term10 = -4
Answer: C
Yes, this solution is not very "mathematical" but the SOLE GOAL with GMAT questions is to answer the question AS ACCURATELY AND AS FAST AS POSSIBLE.
Cheers,
Brent
I like this solution too.
in simple way to understand the logic:-
t1 = 23
tn = -4
total units to be decreased = 23 + 4 = 27
decrement value= 3
no of times the decrement shall run = 27/3 = 9 {23-3= 20-3= 17-3= 14-3= 11-3= 8-3= 5-3= 2-3= -1-3= -4)
tn shall be 9 + 1 = 10
(add 1 becouase of t1 term and rest nine are the no of times the decerement runs)
t1 = 23
tn = -4
total units to be decreased = 23 + 4 = 27
decrement value= 3
no of times the decrement shall run = 27/3 = 9 {23-3= 20-3= 17-3= 14-3= 11-3= 8-3= 5-3= 2-3= -1-3= -4)
tn shall be 9 + 1 = 10
(add 1 becouase of t1 term and rest nine are the no of times the decerement runs)
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