jeni980
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
Answers & Comments
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
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