arithmetic progression

[nisar_cck]

please help me to solve the question.

If in an arithmetic progression sum of m terms divided by sum of n terms
is equal to square of m divided by square of n then show that
m,th term divided by n,th term is equal to 2m- 1 divided by 2n- 1.

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[stapel]

Are we talking about the first "m" terms and the first terms, and of the same progression? So we have:


. . . . .\(\displaystyle \Large{\frac{\frac{m}{2}\mbox{ }\left(a_1\mbox{ }+\mbox{ }a_m\right)}{\frac{n}{2}\mbox{ }\left(a_1\mbox{ }+\mbox{ }a_n\right)}}\mbox{ }=\mbox{ }\frac{m^2}{n^2}\)


. . . . .\(\displaystyle \Large{\frac{m\mbox{ }\left(a_1\mbox{ }+\mbox{ }a_1\mbox{ }+\mbox{ }d(m\mbox{ }-\mbox{ }1)\right)}{n\mbox{ }\left(a_1\mbox{ }+\mbox{ }a_1\mbox{ }+\mbox{ }d(n\mbox{ }-\mbox{ }1)\right)}\mbox{ }=\mbox{ }\frac{m^2}{n^2}}\)


. . . . .\(\displaystyle \Large{\frac{m\mbox{ }\left(2a_1\mbox{ }+\mbox{ }dm\mbox{ }-\mbox{ }d\right)}{n\mbox{ }\left(2a_1\mbox{ }+\mbox{ }dn\mbox{ }-\mbox{ }d\right)}\mbox{ }=\mbox{ }\frac{m^2}{n^2}}\)


...where "d" is the common difference. And we need to show that:


. . . . .\(\displaystyle \Large{\frac{a_m}{a_n}\mbox{ }=\mbox{ }\frac{a_1\mbox{ }+\mbox{ }d(m\mbox{ }-\mbox{ }1)}{a_1\mbox{ }+\mbox{ }d(n\mbox{ }-\mbox{ }1)}\mbox{ }=\mbox{ }\frac{2m\mbox{ }-\mbox{ }1}{2n\mbox{ }-\mbox{ }1}}\)

Is this interpretation correct?

Thank you.

Eliz.

 

[soroban]

Hello, nisar_cck!

I agree with your interpretation, Eliz . . .

\(\displaystyle \text{If in an arithmetic progression: }\;\L\frac{\sum^m_{k=1}a_k}{\sum^n_{k=1}a_k}\;=\;\frac{m^2}{n^2}\)

\(\displaystyle \text{then show that: }\;\L\frac{a_m}{a_n}\:=\:\frac{2m\,-\,1}{2n\,-\,1}\)

\(\displaystyle \L S_m\:=\:\frac{m}{2}[2a\,+\,d(m-1)],\;\;S_n\:=\:\frac{n}{2}[2a\,+\,d(n-1)]\)

Hence, we are given" . \(\displaystyle \L\frac{\frac{m}{2}[2a\,+\,d(m-1)]}{\frac{n}{2}[2a\,+\,(d(n-1)]}\;=\;\frac{m^2}{n^2}\)

. . which simplifies to: .\(\displaystyle 2an\,-\,dn\:=\:2am\,-\,dm\;\;\Rightarrow\;\;d(n\.-\.m)\:=\:2a(n\.-\.m)\)

Assuming \(\displaystyle m\,\neq\,n\), we have: .\(\displaystyle d = 2a\)


Then: .\(\displaystyle a_m\:=\:a\,+\,(m-1)\cdot2a\:=\:a(2m\,-\,1)\)
. . . . . . \(\displaystyle a_n\:=\:a\,+\,(n-1)\cdot2a\;=\:a(2n\,-\,1)\)

Therefore: . \(\displaystyle \L\frac{a_m}{a_n}\:=\:\frac{a(2m\,-\,1)}{a(2n\,-\,1)}\:=\:\frac{2m\,-\,1}{2n\,-\,1}\)