Question

If $a,b,c,d$ are in $H.P.$ then prove that $ab+bc+cd=3ad.$

Answer 1

Given $a,b,c,d$ are in $H.P⟹\frac{1}{a},\frac{1}{b},\frac{1}{c},\frac{1}{d}$ are in $A.P$

let the common difference be $k⟹\frac{1}{b}=\frac{1}{a}+k,\frac{1}{c}=\frac{1}{a}+2k,\frac{1}{d}=\frac{1}{a}+3k$

$ab+bc+cd=abcd(\frac{1}{cd}+\frac{1}{ad}+\frac{1}{ab})=abcd(\frac{1}{cd}+\frac{1}{a}(\frac{1}{b}+\frac{1}{d}))=abcd(\frac{1}{cd}+\frac{2}{ac})$

$=abd(\frac{2}{a}+\frac{1}{d})$

$=abd(\frac{3}{a}+3k)=3ad$