If Sn denotes the sum of first n terms of an A.P. a1 + a2 + a3 + ..., such that Sm/Sn = m^2/n^2 , then am/an = ?

Question

If S_n denotes the sum of first n terms of an A.P. a₁ + a₂ + a₃ + ..., such that \(\frac{S_m}{S_n} = \frac{m^2}{n^2}\), then \(\frac{a_m}{a_n}\) =?

(a) m – 1 : n – 1 

(b) m – n : m + n 

(c) 2m – 1 : 2n – 1 

(d) m + 1 : n + 1

Answer 1

Answer : (c) 2m-1 : 2n-1

\(\frac{S_m}{S_n} = \frac{m/2[2a_1 + (m-1)d]}{n/2[2a_1 +(n-1)d]} = \frac{m^2}{n^2}\) (d → common difference of A.P.)

⇒ n[2a₁ + (m – 1) d] = m[2a₁ + (n – 1) d]

⇒ 2a₁ (n – m) = d (n – m) ⇒ d = 2a₁ 

∴ \(\frac{a_m}{a_n} = \frac{mth \,term}{nth\,term}\) _{= \(\frac{a_1 +(m-1)d}{a_1 + (n-1)d}\)} 

_{=  \(\frac{a_1 +(m-1). 2a_1}{a_1 + (n-1). 2a_1}\)  = \(\frac{ -a_1 +2a_1m}{-a_1 +2a_1n}\)}  

_{= \(\frac{a_1 (2m-1)}{a_1 (2n-1)}\) = \(\frac{(2m-1)}{(2n-1)}\)}