Question

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Answer 1

a : first term of AP

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Answer 2

$\bf{\Huge{\boxed{\sf{\blue{\bf{ANSWER\::}}}}}}$

$\bf{\Large{\underline{\tt{Given\::}}}}}$

If the ratio of sum of the first m & n terms of an A.P. is m² to n².

$\bf{\Large{\underline{\sf{To\:find\::}}}}}$

Show that the ratio of its nth & mth term is (2m-1):(2n-1).

$\bf{\Large{\underline{\sf{\pink{Explanation\::}}}}}}$

We know that formula of the sum of the arithmetic progression:

$\leadsto\sf{\orange{Sn=\frac{n}{2} [2a+(n-1)d]}}$

$\bf{Such\:as}\begin{cases}\sf{a\:=\:First\:term\:of\:an\:A.P.}\\ \sf{d\:=\:Common\:difference.}\\ \sf{Sn\:=\:sum\:of\:n\:terms.}\\ \sf{n\:=\:number\:of\:terms.}\end{cases}}$

According to the question:

$\longmapsto\sf{\frac{Sum\:of\:m\:terms}{Sum\:of\:n\:terms} =\frac{m^{2} }{n^{2} } }$

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

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

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

$\longmapsto\sf{\frac{[2a+(m-1)d]}{[2a+(n-1)d]} =\frac{\cancel{m^{2} }}{\cancel{n^{2} }} *\frac{\cancel{n}}{\cancel{m}} }$

$\longmapsto\sf{\frac{[2a+(m-1)d]}{[2a+(n-1)d]} =\frac{m}{n} ........................(1)}$

We know that nth term of an A.P.;

$\leadsto\sf{\orange{^{a} n=a+(n-1)d}}$

Therefore,

$\longmapsto\sf{\frac{mth\:term}{nth\:term} =\frac{(2m-1)}{(2n-1)} }$

$\longmapsto\sf{\frac{a+(m-1)d}{a+(n-1)d} =\frac{(2m-1)}{(2n-1)} ..................(2)}$

From equation (1), we get;

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

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

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

$\longmapsto\sf{\frac{2a+2(m-1)}{2a+2(m-1)} =\frac{(2m-1)}{(2n-1)} }$

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

$\longmapsto\sf{\frac{\cancel{2}[a+(m-1)d]}{\cancel{2}[a+(n-1)d]} =\frac{(2m-1)}{2n-1)} }$

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

$\longmapsto\sf{\frac{mth\:term\:of\:an\:A.P.}{nth\:term\:of\:a\:A.P.} =\frac{(2m-1)}{(2n-1)} }$

Thus,

$\bf{\large{\boxed{\tt{\green{The\:ratio\:of\:mth\:\&\:nth\:term\:is\:(2m-1):(2n-1).}}}}}$

Proved.