Question

The ratio of the sums of first m and first n of an AP is m²:n². Show that the ratio of its mth and nth terms is (2m-1):(2n-1)​

Answer 1

HOPE THIS ANSWER IS HELPFUL...

Step-by-step explanation:

PLEASE MERA ANSWER KO BRAINLIEST KAR DO PLEASE...

Answer 2

Step-by-step explanation:

Let Sm and Sn be the sum of the first m and first n terms of the AP respectively. Let, a be the first term and d be a common difference.

$\frac{sm}{sn} = \frac{m {}^{2} }{n {}^{2} }$

m/2[2a+(m-1)d]. m

___________. =. n

n/2 [2a+(n-1)d].

2a + (m-1) d. m

__________ = n

2a + (n-1) d

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

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

$md - nd = 2am - 2an$

$(m - n)d \: = 2a(m-n) \\ d = 2a$