If H1,H2,.....Hn be n harmonic means …

We have,a, H1,H2,.......Hn,b in H.P.⇒1a,1H1,1H2,......1Hn,1bThere are n+2 terms⇒1b=1a+(n+2−1)d⇒1b−1a=(n+1)d⇒(n+1)d=a−bab⇒d=(a−b)(n+1)ab1H1=1a+d⇒aH1=1+ad⇒1+aH1=1+1+ad⇒H1+aH1=2+ad ;equation(i)Again1H1=1a+d⇒aH1=1+ad⇒−ad=1−aH1⇒−ad=H1−aH1;equation(ii)equation(ii)÷equation(i), we get,H1−aH1+a=−ad2+ad ;equation(A)Now 1b=1Hn+d⇒1=bHn+bd⇒1−bHn=bd⇒bd=Hn−bHn ;equation(iii)Again1b=1Hn+d⇒1=bHn+bd⇒1+1=1+bHn+bd⇒2−bd=Hn+bHn⇒Hn+bHn=2−bd ;equation(iv)equation(iii)÷equation(iv)⇒Hn−bHn+b=bd2−bd ;equation(B)equation(A)+equation(B)⇒H1−aH1+a+Hn−bHn+b=−ad2+ad+bd2−bd⇒H1−aH1+a+Hn−bHn+b=d(b2−bd−a2+ad)⇒H1−aH1+a+Hn−bHn+b=d⎛⎝⎜⎜b2−b(a−b)(n+1)ab−a2+a(a−b)(n+1)ab⎞⎠⎟⎟=d⎛⎝⎜⎜b2−(a−b)(n+1)a−a2+(a−b)(n+1)b⎞⎠⎟⎟=(a−b)(n+1)ab{ab(n+1)2(n+1)a−(a−b)−ab(n+1)2(n+1)b−(a−b)}=(a−b)ab(n+1)(n+1)ab{12(n+1)a−(a−b)−12(n+1)b−(a−b)}=(a−b)(n+1)ab[2(n+1)b−(a−b)−2(n+1)a+(a−b){2(n+1)a−(a−b)}{2(n+1)b−(a−b)}]=(a−b)(n+1)ab[2(n+1)b−2(n+1)a{2(n+1)a−(a−b)}{2(n+1)b−(a−b)}]=(a−b)ab[2b−2a{2(n+1)a−(a−b)}{2(n+1)b−(a−b)}]