in the arithmetic sequence

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

t(n) = t(n) - 3.

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.

GMATGuruNY wrote:

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

t(n) = t(n) - 3.

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

Brent@GMATPrepNow wrote:

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

Mitch's approach is nice and fast.
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

Thanks 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)

Another solution

Here r=-3
so tn=(n-p)r + tp
And p=1
so tn= -3(n-1) + t1 = -3(n-1) + 23 = -4
so n=10

C.10

Thanks jj Kouakou. Tu es mon frère ivoirien merci bcp