Name: Sum of First n Natural Numbers
Uploaded: 2019-07-19
Duration: 3 min
Description: What are natural numbers? Do you know how to find the sum of the first n natural numbers? Watch this video to get answers to these questions.

Question

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

(2 m - 1) : (2 n - 1)

.

Accepted Answer

Assume $a$ be the first term and $d$ be the common difference of A.P.
Given that, $\frac{s u m o f m t e r m s}{s u m o f n t e r m s} = \frac{m^{2}}{n^{2}}$
Sum of n numbers in A.P. $(S_{n}) = \frac{n}{2} [2 a + (n - 1) d]$ .
Using the above formula we get:
$\Rightarrow \frac{\frac{m}{2} [2 a + (m - 1) d]}{\frac{n}{2} [2 a + (n - 1) d]} = \frac{m^{2}}{n^{2}}$
$\Rightarrow \frac{2 a + (m - 1) d}{2 a + (n - 1) d} = \frac{m}{n} \dots (i)$
Substituting $m by 2 m - 1$ and $n by 2 n - 1$ in $(i)$ , we get
$\frac{2 a + (2 m - 2) d}{2 a + (2 n - 2) d} = \frac{2 m - 1}{2 n - 1}$
$\Rightarrow \frac{a + (m - 1) d}{a + (n - 1) d} = \frac{2 m - 1}{2 n - 1} \dots (i i)$

The general term in A.P. $(a_{n}) = a + (n - 1) d$
Using the above formula we get:
$\frac{m^{t h} t e r m o f A . P .}{n^{t h} t e r m o f A . P .} = \frac{a + (m - 1) d}{a + (n - 1) d} \dots (i i i)$
From $(i i)$ and $(i i i)$ , we get
$\frac{m^{t h} t e r m o f A . P .}{n^{t h} t e r m o f A . P .} = \frac{2 m - 1}{2 n - 1}$
Hence proved.

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 $(2 m - 1) : (2 n - 1)$ .

Important Questions on Sequences and Series

Learn and Prepare for any exam you want !

The ratio of the sums of m and n terms of an A.P. is m2:n2. Show that the ratio of mth and nth term is (2m-1):(2n-1).

Important Questions on Sequences and Series

Learn and Prepare for any exam you want !

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 $(2 m - 1) : (2 n - 1)$ .