If a, b, c, d are in H.P.,then ab + bc + cd i | vedclass

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(A) Since $a, b, c, d$ are in $H.P.$,their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}, \frac{1}{d}$ are in $A.P.$Let the common difference be

Vedclass · Accepted Answer

$3ad$

Vedclass · Answer

(A) Since $a, b, c, d$ are in $H.P.$,their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}, \frac{1}{d}$ are in $A.P.$Let the common difference be $k$. Then $\frac{1}{b} = \frac{1}{a} + k$,$\frac{1}{c} = \frac{1}{a} + 2k$,and $\frac{1}{d} = \frac{1}{a} + 3k$.Alternatively,using the property of $H.P.$,$b = \frac{2ac}{a+c}$ and $c = \frac{2bd}{b+d}$.From $b = \frac{2ac}{a+c}$,we get $\frac{1}{b} = \frac{a+c}{2ac} = \frac{1}{2c} + \frac{1}{2a}$,which implies $2ac = b(a+c) = ab + bc$.Similarly,from $c = \frac{2bd}{b+d}$,we get $2bd = c(b+d) = bc + cd$.Adding these,$ab + 2bc + cd = 2ac + 2bd$.Using the property $b = \frac{2ac}{a+c}$ and $c = \frac{2bd}{b+d}$,we can simplify the expression $ab + bc + cd$.Let $a=1, b=1/2, c=1/3, d=1/4$. Then $ab+bc+cd = (1)(1/2) + (1/2)(1/3) + (1/3)(1/4) = 1/2 + 1/6 + 1/12 = (6+2+1)/12 = 9/12 = 3/4$.Since $3ad = 3(1)(1/4) = 3/4$,the expression is equal to $3ad$.

If $a, b, c, d$ are in $H.P.$,then $ab + bc + cd$ is equal to

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