Question

If the ratio of sum of the first m and n terms of an ap is m2:n2, show that the ratio of its mth and nth terms is (2m-1):(2n-1)?

Answer 1

ans . hope it will help.....

Answer 2

$\Large{\textbf{\underline{\underline{According\;to\;the\;Question}}}}$

Assumption,

First term = a

Common difference = d

Sum of m term = S(m)

Also,

Sum of n term = S(n)

So,

S(m)/S(n) = m²/n²

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

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

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

m{2a + (n - 1)d} = n{2a + (m - 1)d}

d = 2a

Then,

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

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

a(m)/a(n) = 2a + 2ma - 2a/2a + 2na - 2a

a(m)/a(n) = 2ma - a/2na - a

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

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

Therefore,

Proved