If m arithmetic means (A.Ms) and three geometric means (G.Ms) are inserted between 3 and 243 such that 4ᵗʰ A.M. is equal to 2ⁿᵈ G.M., then m is equal to _________ .

Solution

Maths sir · 4 steps

Find the 2nd G.M. first, since it is a fixed number, then make the 4th A.M. land on it.

1. Set up the G.P. Three G.M.s between 3 and 243 make a G.P. of 5 terms, so 243 = 3r⁴. That gives r⁴ = 81 and r² = 9.
2. Second G.M. The 2nd G.M. is the 3rd term of the G.P.: G₂ = 3r² = 3 × 9 = 27. This is true whether r = 3 or r = -3.
3. Set up the A.P. With m A.M.s, the A.P. has m + 2 terms, so 243 is term number m + 2. That means 243 = 3 + (m+1)d, so d = 240/(m+1). The 4th A.M. is the 5th term, A₄ = 3 + 4d.

   3 + 4d = 27 ⇒ d = 6 ⇒ m + 1 = 240/6 = 40

4. So. m + 1 = 40, so m = 39.

The common slip: 40. Using m gaps instead of m + 1, that is d = 240/m, gives m = 40. Between m inserted means and the two ends there are m + 1 steps.

The idea to keep. Inserting n means between a and b gives n + 2 terms, so the step uses n + 1 gaps: d = (b-a)/(n+1) and rⁿ⁺¹ = b/a. The kth mean is term number k + 1.

Answer from the NTA final answer key. Solution by PrepWiser.