Question

plz solve this question.

Answer 1

here's u r answer dear....
let 'a' be the first term and 'd' be the common difference of the AP.
Sm=m/2{ 2a +(m-1)d}
and,
Sn=n/2{ 2a+(n-1)d
now,the terms r given in ratio so,
Sm/Sn=m²/n²
m/2 {2a +(m-1)d
= _______________ =m²/n²

n/2 {2a+(n-1)d
=2a+(m-1)d
__________=m/n
2a+(n-1)d
={2a+(m-1)d}n={2a+(n-1)d}m. (by cross multip....)
=2a(n-m) =d{(n-1)m- (m-1) n}
=2a(n-m)=d(n-m)
=d=2a
now,
Tm/Tn=a+(m-1)d/a+(n-1)d=a+(m-1)2a/a+(n-1)2a=2m-1/2n-1.....hence proved
hope it helps you...
plz mark it as brainliest please...... plz, zzzzz

Answer 2

HELLO DEAR,

given that:-

Sum of m terms of an A.P. = m/2 [2a + (m -1)d]

Sum of n terms of an A.P. = n/2 [2a + (n -1)d]

( m/2 [2a + (m -1)d] ) / (n/2 [2a + (n -1)d] ) = m² : n²

=> [2a + md - d] / [2a + nd - d] = m/n

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

=> 2an - 2am = nd - md

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

=> 2a = d----------(1)

Ratio of m th term to n th term:

[a + (m - 1)d] / [a + (n - 1)d]

=> [a + (m - 1)2a] / [a + (n - 1)2a]---------using(1)

=>a[1 + 2m - 2] / a[1 + 2n-2]

=>(2m - 1)/(2n -1)

I HOPE ITS HELP YOU DEAR,
THANKS