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Explanation

Let a and b be two numbers and `A_(1),A_(2),A_(3),…A_(n)` be n A.M's between a and b.
Then `a,A_(1),A_(2),…,A_(n)`,b are in A.P.
There are (n+2) terms in the series
`rArra+(n+1)d=b`
`rArrd=(b-1)/(n+1)`
`rArrA_(1)=p=a+(b-1)/(n+1)=(an+b)/(n+1)` ..(1)
The first H.M between a and b, when nHM's are inserted between a and b can be obtained by replacing a by `1/a` and b by `1/b` in eq. (1) and then taking its reciprocal.
Therefore,`q=1/(((1/a)n+1/b)/(n+1))=((n+1)ab)/(bn+a)`
Substitute b=p(n+1)-an[from (1)] in equation (2) to get
`aq+nq[p(n+1)-an]=(n+1)a[p(n+1)-an]`
`rArrna^(2)-[(n+1)p+(n-1)q]a+npq=0`
`rArr " Discriminant " ge0 (because ` is a real)
`rArr[(n+1)p+(n-1)q]^(2)-4n^(2)pqge0`
`rArr(n-1)^(2)q^(2)+{2(n^(2)-1)-4n^(2)}pq+(n+1)^(2)p^(2)ge0`
`rArrq^(2)-2(n^(2)+1)/((n-1)^(2))pq+((n+1)/(n-1))^(2)p^(2)ge0`
`rArr[q-p((n+1)/(n-1))^(2)][q-p]ge0`
`rArrq` can not lie between p and q `((n+1)/(n-1))^(2)`
Also `[p((n+1)/(n-1))^(2)-q][p-q]ge0`
`rArr[p-q((n-1)/(n+1))^(2)][p-q]ge0`
`rArr p` can not lie between q `((n-1)/(n+1))^(2)` and q