If a1,a2,…an are in H.P. then the expression a1a2+a2a3+…+an−1an is equal to

a1,a2,…an are in H.P.⇒ 1a1,1a2,…1an are in A.P.⇒ 1a2−1a1=1a3−1a2=⋯1an−1an−1=d(say)⇒ a1a2=1da1−a2,a2a3=1da2−a3,…an−1an=1dan−1−anThus,   a1a2+a2a3+…+an−1an                         =1da1−a2+a2−a3+…+an−1−an              =1da1−anBut    1an=1a1+(n−1)d⇒a1−anana1=(n−1)d∴ a1a2+a2a3+…+an−1an=(n−1)a1an