Math Is Fun Forum

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: What am I doing wrong? #Arithmetic_Progressions

On second thought, it is her answer and she should explain it. I will have her do it. I am tired.

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

ElainaVW

Member

Registered: 2013-04-29

Posts: 580

Re: What am I doing wrong? #Arithmetic_Progressions

For n = 1...3

Let

The equations are

I used the arithmetic nth term formula.

From what is above.

Last edited by ElainaVW (2014-03-09 22:52:34)

gourish

Member

Registered: 2013-05-28

Posts: 153

Re: What am I doing wrong? #Arithmetic_Progressions

sorry to jump in but i do think that the ratio is for the sum of n terms of two series just solve it like
3n+4/2n+7=sum of n terms of Ist a.p/sum of n terms of 2nd a.p.
3n+4/2n+7=n/2*{2a+(n-1)d)/n/2*{2b+(n-1)c).....-(i)
where a and b are first terms of 1st a.p and 2nd a.p and d and c are common difference of 1st and 2nd a.p
solving that and putting n-1=2(m-1). n=2m+1 and hence using this value of 'n' the R.H.S. of eq. (i) becomes the ratio of mth terms of 1st and 2nd terms while the L.H.S. gives its value in terms of 'm' i had this problem in my higher secondary exams and this was the right answer

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"The man was just too bored so he invented maths for fun"
-some wise guy

gourish

Member

Registered: 2013-05-28

Posts: 153

Re: What am I doing wrong? #Arithmetic_Progressions

sorry it's actually n=2m-1 and the ratio comes out to be 6m+1/4m+5 which is the same as got by Elaina but much faster and simpler.. the idea was to get the m'th rem of the A.P. but taking n-1=2(m-1) we get that term on both numerator and denominator and cancel out the '2' which is common to both of them

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"The man was just too bored so he invented maths for fun"
-some wise guy