Question

PLEASE ANSWER IT AS AN ATTACHMENT​

Answer 1

Answer:

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Answer 2

As we know that : sum of m and n terms will be respectively :

$\frac{m}{2} (2a + (m - 1)d) \: and \: \frac{n}{2} (2a + (n - 1)d)$

According to the question :

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

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

On putting m = 2m-1 and n = 2n-1

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

HENCE PROVED ✔️✔️

✌️_____ FÓLLÒW MË _____✌️