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 $ab + bc = 2ac$. From $c = \frac{2bd}{b+d}$,we get $bc + cd = 2bd$.Using the property for four terms in $H.P.$,we have $ab + bc + cd = 3ad$.Verification: Let $a=1, b=\frac{1}{2}, c=\frac{1}{3}, d=\frac{1}{4}$.$ab + bc + cd = (1 	imes \frac{1}{2}) + (\frac{1}{2} 	imes \frac{1}{3}) + (\frac{1}{3} 	imes \frac{1}{4}) = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} = \frac{6+2+1}{12} = \frac{9}{12} = \frac{3}{4}$.Since $3ad = 3(1 	imes \frac{1}{4}) = \frac{3}{4}$,the result is $3ad$.

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

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