If the ratio of sum of the first m and n terms of an AP is m2 : n2. . .

Problem : If the ratio of sum of first 'm' terms and 'n' terms of an AP is m²:n², show that the ratio of it's mth and nth terms is (2m-1):(2n-1)

Doubt by Muskan, Gauri & Kanika

Solution : 

We know, 
S_n=n/2[2a+(n-1)d]

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

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

2an+(m-1)nd=2am+(n-1)md
(m-1)nd-(n-1)md=2am-2an
d[mn-n-mn+m]=2a(m-n)
d[m-n]=2a(m-n)
d=2a — (1)

We also know that, 
a_n=a+(n-1)d

a_m/a_n=[a+(m-1)d]/[a+(n-1)d]
a_m/a_n=[a+(m-1)2a]/[a+(n-1)2a] [∵d=2a]
a_m/a_n=[a+2am-2a]/[a+2an-2a] 
a_m/a_n=[2am-a]/[2an-a]
a_m/a_n=a[2m-1]/a[2n-1]
a_m/a_n=[2m-1]/[2n-1]
a_m:a_n=(2m-1):(2n-1)