Question

the ratio of sum of m and n terms of an ap is m^2:n^2 show that mth term and nth term is 2m-1 : 2n-1​

Answer 1

$\huge\bold{\gray{\sf{Answer:}}}$

Explanation:

Let sn and sm be the sum of first m and n terms of an AP respectively.a be the first term and d be the common difference.

According to the question:

Formula used here :

$\bold{\boxed{Sn = \frac{n}{2} [2a + (n - 1)d]}}$

$\bold{\boxed{An = a + (n - 1)d}}$

$\frac{Sm}{Sn} = \frac{ {m}^{2} }{ {n}^{2} }$

$= > \frac{ \frac{m}{2} }{ \frac{n}{2} } = \frac{[2a + (m - 1)d]}{[2a + (n - 1)d]} = \frac{ {m}^{2} }{ {n}^{2} }$

$= > \frac{2a + (m - 1) d}{2a + (n - 1)d} = \frac{m}{n}$

$= > n(2a + (m - 1)d) = m(2a + (n - 1)d$

$= > 2an + mnd - nd + 2am + mnd - nd$

$= > md - nd = 2am - 2an$

$= > (m - 1)d = 2a(m - n)$

$\bold{\red{d = 2a}}$

Given that ratio of mth term and nth term is :

$= > \frac{am}{an} = \frac{a + (m - 1)d}{a + (n - 1)d} = \frac{a + (m - 1){\red{2a}}}{a + (n - 1){\red{2a}}}$

$= \frac{a (1 + 2m - 2)}{a(1 + 2n - 2)} = \frac{2m - 1}{2n - 1}$

Answer 2

Answer:

answer for the given problem is given