If the ratio of the sum of $m$ terms and $n$ terms of an $A . P .$ be $m^{2} : n^{2}$ , prove that the ratio of its $m t h$ and nth terms will be $2 m - 1 : 2 n - 1$ .

Open in App

Solution

$\frac{S_{m}}{S_{n}} = \frac{m^{2}}{n^{2}} = k$

$∴ \frac{S_{m}}{m^{2}} = \frac{3 n}{n^{2}} = k$

$S_{m} = m^{2} k, S_{n} = n^{2} k$

Now, $\frac{T_{m}}{T_{n}} = \frac{S_{m} - S_{m} - 1}{S_{n} - S_{n} - 1} = \frac{k (m^{2} - (m - 1)^{2})}{k (n^{2} - (n - 1)^{2})}$

$= \frac{2 m - 1}{2 n - 1}$

Suggest Corrections

Similar questions

If the ratio of the sum m terms and n terms of an A.P. are m²:n² . Prove that the ratio of its m^th and n^th terms will be (2m-1):(2n-1).

If the ratio of sum of m terms and n terms of an A.P. is $m^{2}$ : $n^{2}$ , then the ratio of its $m^{t h}$ and $n^{t h}$ terms will be

If the ratio of sum of m terms and n terms of an A.P. is $m^{2}$ : $n^{2}$ , then the ratio of its $m^{t h}$ and $n^{t h}$ terms will be :

Q. The ratio of the sums of m and n terms of an A.P. is

m^{2} : n^{2}

. Show that the ratio of

m^{t h}

and

n^{t h}

term is (2m - 1) : (2n - 1).

The ratio of the sums of m and n terms of an A.P. is $m^{2}$ : $n^{2}$ . Show that the ratio of the $m^{t h}$ and $n^{t h}$ term is $(2 m - 1) : (2 n - 1)$ . [4 MARKS]

Join BYJU'S Learning Program

If the ratio of the sum of m terms and n terms of an A.P. be m2:n2, prove that the ratio of its mth and nth terms will be 2m−1:2n−1.

SmSn=m2n2=k ∴Smm2=3nn2=k Sm=m2k,Sn=n2k Now, TmTn=Sm−Sm−1Sn−Sn−1=k(m2−(m−1)2)k(n2−(n−1)2) =2m−12n−1

If the ratio of the sum of $m$ terms and $n$ terms of an $A . P .$ be $m^{2} : n^{2}$ , prove that the ratio of its $m t h$ and nth terms will be $2 m - 1 : 2 n - 1$ .

$\frac{S_{m}}{S_{n}} = \frac{m^{2}}{n^{2}} = k$

$∴ \frac{S_{m}}{m^{2}} = \frac{3 n}{n^{2}} = k$

$S_{m} = m^{2} k, S_{n} = n^{2} k$

Now, $\frac{T_{m}}{T_{n}} = \frac{S_{m} - S_{m} - 1}{S_{n} - S_{n} - 1} = \frac{k (m^{2} - (m - 1)^{2})}{k (n^{2} - (n - 1)^{2})}$

$= \frac{2 m - 1}{2 n - 1}$