Question

This is Arithmetic progression
pls help me pls pls pls​

Answer 1

Here's your answer..... :)

Answer 2

Answer:

Okay so

according to question

$\frac{sm}{sn} = \frac{ \frac{m}{2} \times (2a + (n - 1)d )}{ \frac{n}{2} \times 2a + (n - 1)d)} = \frac{m {}^{2} }{n {}^{2} } \\ now \: 2 \: is \: cancelled \: \\ and \: m \: and \: n \: are \: also \: cancelled \: \\ so \: \frac{ (2a + (m - 1)d )}{ 2a + (n - 1)d)} = \frac{m}{n} \\ \\ \\ cros s \: multiply \: \\ \\ n(2a + (m - 1)d ) \: = m(2a + (n - 1)d ) \\ \\ 2an + dn(m - 1) = 2am + \: dm(n - 1) \\ 2an - 2am + dn( m- 1) - dm(n - 1) \\ 2a(n - m) + d(mn - n - mn \: + m) \\ 2a(n - m) + d(m - n) \\ 2a(n - m) - d(m - n) = 0 \\ 2a - d = 0 \\ 2a = d$

now 2a = d

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

Hence proved.