Question

ratio of the sum of m terms sm/sn=m^2/n^2 show that ratio of mth term am/an=2m-1/2n-1

Answer 1

Answer:

Step-by-step explanation:

Given ratio of the sum of m terms sm/sn=m^2/n^2 show that ratio of mth term am/an=2m-1/2n-1

We know that sum of n terms of an A.P.

Sn = n/2 (2a + (n – 1) d)

Similarly sum of m terms will be

Sm = m/2 (2a + (m – 1)d)

Given ratio  

Sm / Sn = m^2 / n^2

m/2 (2a + (m – 1)d) / n/2 (2a + (n – 1)d) = m^2 / n^2

m/n (2a + (m – 1)d /2a + (n – 1)d) = m^2 / n^2

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

We know that a n = a + (n -1) d

         and a m = a + (m – 1)d ( because m th and n th term)

So m th term / n th term = (2m – 1) / (2n – 1)

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

From 1  

Replacing m with 2m – 1 and n with 2n – 1

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

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

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

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

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

m th term of A.P / n th term of A.P = (2m – 1) / (2n – 1)

So ratio of  am / a n = (2m – 1) / (2n – 1)