Question

Dum h to 19no. Ka answer do .



I challenge you.

Answer 1

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

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

now put m=2n-1 & n=2m-1, We get

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

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

then

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

therefore

tm/tn = (2n-1)/(2m-1)

[since {a+(m-1)d}=tm & {a+(n-1)d}=tn]

hence proved