Let a1, a2, a3, ...... be the terms of an A.P. If [(a1+a2+a3+...+ap)/(a1+a2+a3+...+aq)] = p^2/q^2 (p ≠ q), then find a6/a21.

Question

Let a₁, a₂, a₃, ...... be the terms of an A.P. If \(\frac{a_1 +a_2 +a_3 +...+a_p}{a_1 + a_2 + a_3 + ... +a_q}\) = \(\frac{p^2}{q^2}\) (p ≠ q), then find \(\frac{a_6}{a_{21}}\).

Answer 1

Let d be the common difference for the A.P.; a₁, a₂, a₃, .... 

Then, \(\frac{S_p}{S_q}\) = \(\frac{p/2[2a_1 + (p-1) d ]}{q/2 [ 2a_1 + (q - 1) d]}\) = \(\frac{p^2}{q^2}\) 

⇒ \(\frac{2a_1 + (p-1) d }{ 2a_1 + (q - 1) d}\) = \(\frac{p}{q}\)

⇒ q [2a₁ + (p – 1) d] = p [2a₁ + (q –1) d] 

⇒ 2a₁q +pqd - qd = 2a₁p + pqd - pd 

⇒ pd – qd = 2a₁p – 2a₁q 

⇒ d (p – q) = 2a₁ (p – q) 

⇒ d = 2a₁

Now