if the ratio of sum of the first m and n terms of an A.P is m² : n², show that the ratio of its mth and nth terms is (2m-1) : (2n-1)

SARAHann
Mar 25, 2017

#1**+1 **

First member of the A.P- a

a_{n}-a_{n-1}=constant=d

You shouldnt have a problem if you know how to sum an A.P

Im starting to think you post every math problem you have HERE without even solving it.

Ehrlich
Mar 25, 2017

#4**+2 **

If i don't understand the question is there any point in trying to solve it?

SARAHann
Mar 25, 2017

#5**+2 **

And how can YOU say about that?And i do solve them. part of posting them here is to check wheather i did it correctly.pls don't think things like that. it does'nt help.

SARAHann
Mar 25, 2017

#7**0 **

Yes there is a point... its fine to not understand it on the first try. And im sure you understood the question itself. English is not my first language and i understood what the meaning of the question is.

Besides, theres no point at asking for the solution of a question of the form "prove that if ___ then ___" because there is only one final answer. If you got a solution then all you need to do is check the way you got there is legit, not compare with other people.

The point of such questions is to get the information you know to a simple form, helping you to a simple form.

Ill give you a hint: you can prove that d=2a OR n=m. Both of these can lead you to the answer you need.

Ehrlich
Mar 25, 2017

#8**+2 **

I DID'NT ask any questions of the form prove that if. and i do know that prove that if questions has only 1 answer.

SARAHann
Mar 25, 2017

#10**+2 **

Hi Sara,

It is always hard for answerers to get a nice compromise between (a) doing all of someones homework and (b) causing them massively more frustration by not answering with enough detail.

I can see our guest has caused you a lot of frustration and I am sorry about that.

BUT

I do agree with guest that people who post questions, not just you, everyone, should try to show that they have had a shot at doing it by themselves. Also I do not think that full answers should very often be given especially if the asker has just done a dump and run. I am NOT refering to you here. You did NOT do a dump and run which is why I have come back and answered your question in full :)

If you do not understand then let me know :)

Melody
Mar 26, 2017

#2**0 **

Use the fact that the ratio of the two sums is m squared/ n squared to derive a relationship between a and d. It's easy after that.

Guest Mar 25, 2017

#9**+1 **

Hi Sara,

if the ratio of sum of the first m and n terms of an A.P is m² : n², show that the ratio of its mth and nth terms is (2m-1) : (2n-1)

Let's take a look.

The sum of the first m terms of an AP is \(S_m=\frac{m}{2}[2a+(m-1)d]\)

The sum of the first n terms of an AP is \(S_n=\frac{n}{2}[2a+(n-1)d]\)

The mth term is \(T_m=a+(m-1)d\)

The nth term is \(T_n=a+(n-1)d\)

So given:

\(\frac{\frac{m}{2}[2a+(m-1)d]}{\frac{n}{2}[2a+(n-1)d]}=\frac{m^2}{n^2} \)

SHOW THAT:

\(\frac{a+(m-1)d}{a+(n-1)d}=\frac{2m-1}{2n-1}\)

Ok lets look at what we know is true:

\(\frac{\frac{m}{2}[2a+(m-1)d]}{\frac{n}{2}[2a+(n-1)d]}=\frac{m^2}{n^2}\\~\\ \frac{[2a+(m-1)d]}{[2a+(n-1)d]}=\frac{m}{n}\\~\\ n[2a+dm-d]=m[2a+dn-d]\\ 2an+dmn-dn=2am+dnm-dm\\ 2an-2am=dnm-dm-dmn+dn\\ 2a(n-m)=d(n-m)\\ d=2a\\ \)

Now I need to show the following

\(\frac{a+(m-1)d}{a+(n-1)d}=\frac{2m-1}{2n-1}\)

So I am going to sube d=2a into the LHS

\(LHS\\=\frac{a+(m-1)*2a}{a+(n-1)*2a}\\ =\frac{a+2am-2a}{a+2an-2a}\\ =\frac{2am-a}{2an-a}\\ =\frac{a(2m-1)}{a(2n-1)}\\ =\frac{2m-1}{2n-1}\\ =RHS\)

there you go, that is how it is done :))

Melody
Mar 26, 2017