Question

the ratio of the m and n terms of an AP is m2:n2 show that the ratio of the mth and nth terms is (2m-1):(2n-1)​

Answer 1

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

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

Sm/Sn=m2/n2

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

then in divide m , n, 2cancel out

then we will get

2a+(m-1) d=m/2a+(n-1) d=n

then cross multiplication

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

then mnd cancel

md-nd=2am-2an

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

then (m-n) is cancel

d=2a

Am/An=a+(m-1) d/a+(n-1) d

put the value d=2a in above equation

=a+(m-1) 2a/a+(n-1)2a

= a+2am-2a/a+2an-2a

=2am-a/2an-a

then taking a common then cancel out

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

a is cancel

= 2m-1:2n-1

Answer 2

Your answer is in above of the attachment............