Question

If n A.M.s are inserted between 20 and 80 such that first mean:last mean =1:3. Find n

Answer 1

Let (A1, A2, A3, …...An), be n arithmetic means between 20 and 80 and let d be the common difference between the terms of the A.P.

Then, d = (b – a)/(n + 1) [ where a and b are the 1st and last term respectively]
= (80 – 20)/(n + 1)
= 60 / (n + 1)

A1 = a + d = 20 + d
= 20 + 60/(n + 1)
= 20{1 + 3/(n+1)}
= 20{(n + 4)/(n + 1)}

An = b - d = 80 - d
= 80 - 60/(n + 1)
= 20{4 - 3(n+1)}
= 20 [(4n + 1)/(n + 1)]

Now A1/An = 1/3 [given]

⇒ 20[(n + 4)/(n + 1)] / 20 [(4n + 1) / (n + 1)] = 1/3

⇒ (n + 4) / (4n + 1) = 1/3

⇒ 4n + 1 = 3n + 12

n = 11. Ans.