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[" If "a(1),a(2),...a(n)" are in H.P.,th...

[" If "a_(1),a_(2),...a_(n)" are in H.P.,then the expression "a_(1)a_(2)],[+a_(2)a_(3)+...+a_(n-1)a_(n)" is equal to "],[[" (a) "(n-1)a_(1)a_(n)," (b) "n(a_(1)-a_(n))],[" (c) "(n-1)(a_(1)-a_(n))," Md ")na_(1)a_(n)]]

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