Question

The ratio of sums of m and n terms of an A.P is \[{{m}^{2}}:{{n}^{2}}\] . Show that the ratio of the ${m}^{th}$ and ${n}^{th}$ term is \[(2m-1):(2n-1)\] .

Answer 1

Hint: Given the ratio of sums of m and n terms of an A.P is \[{{m}^{2}}:{{n}^{2}}\]. We have to show that the ratio of the ${m}^{th}$ and ${n}^{th}$ term is \[(2m-1):(2n-1)\]. Solving the given ratio with the sum of n terms formula and then substitute the obtained value in the ${n}^{th}$ term.
Complete step-by-step answer:
Sum of m terms in A,P is given by \[{{s}_{m}}=\dfrac{m}{2}\left( 2a+\left( m-1 \right)d \right)\]
Sum of n terms in A,P is given by \[{{s}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)\]
The given question is \[\dfrac{{{s}_{m}}}{{{s}_{n}}}=\dfrac{{{m}^{2}}}{{{n}^{2}}}\]
By writing the formulas in the above equation we get the further equation as \[\dfrac{\dfrac{m}{2}\left( 2a+\left( m-1 \right)d \right)}{\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)}=\dfrac{{{m}^{2}}}{{{n}^{2}}}\]
Now cancelling all the terms of common we get the further equation as \[2an+mnd-nd=2am+mnd-md\]
Now again cancelling all the terms of common we get the further equation as \[2a(n-m)=d(n-m)\]
By further simplifying we get the equation as \[d=2a\]
The ${n}^{th}$ term in A.P is \[a+\left( n-1 \right)d\]
Now we have to find the ratio of ${m}^{th}$ term and ${n}^{th}$ term,
Writing the formulas as shown, \[\dfrac{a+\left( m-1 \right)d}{a+\left( n-1 \right)d}\]
Now substitute the value of \[d=2a\] in the above equation we get, \[\dfrac{a+(m-1)2a}{a+\left( n-1 \right)2a}\]
Further simplifying the equation the equation is as \[\dfrac{a+2am-2a}{a+2an-2a}\]
Now taking a as common and writing the further is shown below: \[\dfrac{a(2m-1)}{a(2n-1)}\]
Cancelling the ‘a’ term, we get the solution as \[\dfrac{(2m-1)}{(2n-1)}\]
Hence showed that the ratio of the ${m}^{th}$ and ${n}^{th}$ term is \[(2m-1):(2n-1)\] .
Note: Write all the formulas and then solve all the equations carefully. Alternatively, If this question is given as MCQ then solve this by taking a small series like 1+2+3+4 and substitute the values and get the desired solution.