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# Maths3

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Manager
Joined: 29 Apr 2003
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10 Aug 2003, 10:02
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

I have forgotton basic Maths.

Could anyone enlighten on the formula for nth term
as attached?

[/code]

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10 Aug 2003, 18:38
Don't use a formula. Look at what a2 and a3 are. Maybe a4. Generalize for n.
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

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Manager
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18 Aug 2003, 09:24
I remember I did it in "Sec. 3" when I was 15 years.
I had not done this arithmetic maths since then.

If n = 1, a2 = a1+4 = 5
If n = 2. a3 = a2+4 = 9
If n = 3, a4 = a3+4 = 13
If n = n, an+1 = an+4 = ?

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18 Aug 2003, 10:00
looks like (4n - 3) to me.

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18 Aug 2003, 13:13

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Intern
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18 Aug 2003, 15:28

Here is the solution

Arithematic Progression
an=a+(n-1)d

Here a=1 and d=4 substitute the values in the equation to get the answer which is 4n-3

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20 Jan 2004, 22:48
I believe it should be 4n+1

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21 Jan 2004, 04:23
a1=1, a2=5, a3=9,........ so an= 4n-3

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21 Jan 2004, 06:37
It should be 4n-3

if n = 0 then 4n+1 yeilds 1 which is incorrect
if n = 1 then 4n-3 yeilds 1

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21 Jan 2004, 06:37
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# Maths3

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