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GMAT Prep - Arithmetic Sequence

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by sangeethai » Thu Feb 21, 2008 4:47 pm
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

1) -1
2) 7
3) 10
4) 14
5) 20

Correct ans is 3) 10

Please shed some light on this problem. Thanks.
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by camitava » Thu Feb 21, 2008 6:12 pm
sangeethai,
u can take tn = -4 as the first term of the series. now use the formula -
tn = a + (n - 1) * d
where tn = -4, a = -4 and d = 3 , u can get n = 10.
Correct me If I am wrong


Regards,

Amitava
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by siddarthd2919 » Thu Feb 21, 2008 7:19 pm
hi camitava,

the formula u gave is wrong. the formula is

an=a1+(n-1)*d

we have an=23,a1=-4

23 = -4 +(n-1)*3
23 = -4 +3n-3
3n = 30

n= 10
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by camitava » Fri Feb 22, 2008 4:33 am
siddarthd2919,
if u check my post again, u will find that i have taken -4 as the first term and 3 as the common diff. So this comes same if u take 23 as first term, -4 as nth term and -3 as common diff.
Te main thing is the formula - tn = a + (n - 1) * d - now the way u will use this formula depends on u.
Correct me If I am wrong


Regards,

Amitava
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Thanks - Knowing the formula really help

by sangeethai » Fri Feb 22, 2008 6:58 am
Hi Amitava and Siddharth,
Thanks for your help. I have learned to solve this problem.

regards,
Sangeetha
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by khurram » Fri Feb 22, 2008 10:23 am
hello

What heading would this sort of formula be found under. Did not know such a formula existed so want to revise it.

Also, how did you guys calculate d as 3.
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by camitava » Fri Feb 22, 2008 10:28 am
khurram, d is he common difference. as te question says tn = tn-1 - 3. so the common diff is 3 or -3 - the way u will take. d or common diff means the diff between two consecutive terms of the series (AP - Arithmetic Progression). u will get more input f u do some google search on arithmetic progression.
Correct me If I am wrong


Regards,

Amitava
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by khurram » Fri Feb 22, 2008 12:00 pm
Thanks will do tonight

khurram
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by disccomp » Mon Aug 08, 2011 12:35 pm
I was trying to wrap my mind around the sequential number formula, When I realized the problem is a simple linear function, just use the slope intercept formula:
y = mx + b

x => n
y => an
m(slope) => an-1 (The difference between two consecutive numbers in a sequence)
b(y-intercept) =>c

So for the given problem:
23 = -3(1) + b, solve for b: b=26

-4 = -3n + 26, subtract 26 from both sides
-30 = -3n, solve for n: n = 10

No need to memorize a separate formula.
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by edvhou812 » Wed Sep 28, 2011 10:41 pm
disccomp wrote:I was trying to wrap my mind around the sequential number formula, When I realized the problem is a simple linear function, just use the slope intercept formula:
y = mx + b

x => n
y => an
m(slope) => an-1 (The difference between two consecutive numbers in a sequence)
b(y-intercept) =>c

So for the given problem:
23 = -3(1) + b, solve for b: b=26

-4 = -3n + 26, subtract 26 from both sides
-30 = -3n, solve for n: n = 10

No need to memorize a separate formula.
Keep it simple, stupid. I'd thank this post 10 times if I could.
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by studentps2011 » Thu Sep 29, 2011 9:55 am
disccomp wrote:I was trying to wrap my mind around the sequential number formula, When I realized the problem is a simple linear function, just use the slope intercept formula:
y = mx + b

x => n
y => an
m(slope) => an-1 (The difference between two consecutive numbers in a sequence)
b(y-intercept) =>c

So for the given problem:
23 = -3(1) + b, solve for b: b=26

-4 = -3n + 26, subtract 26 from both sides
-30 = -3n, solve for n: n = 10

No need to memorize a separate formula.
Wow! that's a cool way to look at this problem
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