# 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 \[(2m-1):(2n-1)\].

Answer

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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.

Complete step-by-step answer:

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.

Complete step-by-step answer:

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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