Let two numbers be a and b and `A_(1), A_(2),..., A_(n)` be n arithmetic means between a and b. Then, `a, A_(1), A_(2),...,A_(n)`, b are in AP with common difference
`d = (b-a)/(n +1)`
`:.p = A_(1) = a + d = a + (b-a)/(n +1)`
`rArr p = (na + b)/(n+1)`...(i) ltbr. Let `H_(1), H_(2),...,H_(n)` be n harmonic means between a and b.
`:. (1)/(a), (1)/(H_(1)), (1)/(H_(2)),..., (1)/(H_(n)), (1)/(b)` is an AP with common difference, `D = ((a-b))/((n+1) ab)`
`:. (1)/(q) = (1)/(a) + D rArr (1)/(q) = (1)/(a) + ((a -b))/((n +1) ab)`
`rArr (1)/(q) =(nb +a)/((n+1) ab)`
`rArr q = ((n +1) ab)/(nb + a)`...(ii)
From Eq. (i)
`b = (n +1)p - na`
Putting it in Eq. (ii), we get
`q{n (n +1) p -n^(2) a + a} = (n + 1) a {(n +1) p - na}`
`rArr n (n + 1) a^(2) - {(n +1)^(2) p + (n^(2) - 1) q}a + n (n +1) pq = 0`
`rArr na^(2) - {(n +1) p +(n-1) q} a + npq = 0`
Since, a is real, therefore
`{(n +1) p + (n -1)q}^(2) - 4n^(2) pq gt 0`
`rArr (n +1)^(2) p^(2) + (n-1)^(2) q^(2) + 2(n^(2) -1) pq - 4n^(2) pq gt 0`
`rArr (n +1)^(2) p^(2) + (n-1)^(2) q^(2) -2 (n^(2) + 1) pq gt 0`
`rArr q^(2) - (2n (n^(2) + 1))/((n -1)^(2)) pq + ((n +1)/(n-1))^(2) p^(2) gt 0`
`rArr q^(2) - {1 + ((n+1)/(n-1))^(2)}pq + ((n+1)/(n-1))^(2) p^(2) gt 0`
`rArr (q-p) {a - ((n+1)/(n-1))^(2)p} gt 0`
`rArr q lt p or q gt ((n+1)/(n-1))^(2) p`
`:' {((n +1)/(n-1))^(2) p gt q}`
Hence, q cannot lie between p and `((n +1)/(n-1))^(2)p`
