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# The ratio of the sums of first $m$ and first $n$ terms of an arithmetic series is ${{m}^{2}}:{{n}^{2}}$. Show that the ratio of the ${{m}^{th}}$ and ${{n}^{th}}$ terms is $(2m-1):(2n-1)$. Verified
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Hint: Use the formula of ${{r}^{th}}$ term and sum of first r terms of an arithmetic series. Apply it for m and n and calculate the ratio from the equations.

Let the first term and common difference of the arithmetic series is $a$ and $d$ respectively.
We know that the sum of first r terms of an arithmetic series is $\dfrac{r}{2}[2a+(r-1)d]$ as from the formulas of arithmetic series.

So, according to the question, $\dfrac{\dfrac{m}{2}[2a+(m-1)d]}{\dfrac{n}{2}[2a+(n-1)d]}=\dfrac{{{m}^{2}}}{{{n}^{2}}}$
$\Rightarrow$ $\dfrac{2a+(m-1)d}{2a+(n-1)d}=\dfrac{m}{n}$
$\Rightarrow$ $2an+n(m-1)d=2am+m(n-1)d$
$\Rightarrow$ $d[n(m-1)-m(n-1)]=2am-2an$
$\Rightarrow$ $d(m-n)=2a(m-n)$
$\Rightarrow$ $d=2a$ [As $m\ne n$ ]…………… (1)

Now, we know that the ${{r}^{th}}$ term of an arithmetic series is $a+(r-1)d$ where a is the first term and d is the common difference of the arithmetic series.

So, ratio of the ${{m}^{th}}$ and ${{n}^{th}}$ terms is, $\dfrac{a+(m-1)d}{a+(n-1)d}$ ……………… (2)
From (1) putting the value of d in (2) we get, $\dfrac{a+(m-1)d}{a+(n-1)d}$ = $\dfrac{a+2a(m-1)}{a+2a(n-1)}$
= $\dfrac{a+2am-2a}{a+2an-2a}$=$\dfrac{2am-a}{2an-a}=\dfrac{2m-1}{2n-1}$

Hence, the ratio of the ${{m}^{th}}$ and ${{n}^{th}}$ terms is $(2m-1):(2n-1)$.

Note: This problem uses basic properties and formulas of an arithmetic series. While cancelling out m and n from the equation we have assumed that m is not equal to n as if they were the same, then it would be a trivial one. In addition, we have assumed that a is nonzero while cancelling out because if a is zero then d will also be zero which will contradict it to be an arithmetic series.
Last updated date: 26th Sep 2023
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