Let two numbers be a and b and A1​,A2​,…,An​ be n arithmetic means between a and b. Then, a,A1​,A2​,…,An​,b are in AP with common differenced=n+1b−a​⇒p=A1​=a+d=a+n+1b−a​⇒p=n+1na+b​Let H1​,H2​,…,Hn​ be n harmonic means between a and b.∴a1​,H1​1​,H3​1​,…,Hn​1​,b1​ is an AP with commondifference, D=(n+1)ab(a−b)​.∴⇒​q1​=a1​+D⇒q1​=a1​+(n+1)ab(a−b)​q1​=(n+1)abnb+a​⇒q=nb+a(n+1)ab​…​From Eq. (i), b=(n+1)p−na.Putting it in Eq. (ii), we get⇒​q{n(n+1)p−n2a+a}=(n+1)a{(n+1)p−na}n(n+1)a2−{(n+1)2p+(n2−1)q}a+n(n+1)pq=0​⇒na2−{(n+1)p+(n−1)q}a+npq=0Since, a is real, therefore⇒⇒⇒​{(n+1)p+(n−1)q}2−4n2pq>0(n+1)2p2+(n−1)2q2+2(n2−1)pq−4n2pq>0(n+1)2p2+(n−1)2q2−2(n2+1)pqq2−(n−1)22(n2+1)​pq+(n−1n+1​)2p2>0​>0​⇒⇒​q2−{1+(n−1n+1​)2}pq+(n−1n+1)​)2p2>0(q−p){q−(n−1n+1​)2p}>0​⇒∵​q(n−1n+1​)2p{(n−1n+1​)2p>p}​Hence, q cannot lie between p and (n−1n+1​)2p.