Question

If $a.b,c,d$ are in H.P., then the value of $ab+bc+cd$ is equals

• A. $3ac$
• B. $3ad$
• C. $(a+b)(c+d)$
• D. $a+b+c+d$

Answer 1

The correct option is B $3ad$

Given, $a.b,c,d$ are in H.P

So $\frac{1}{a},\frac{1}{b},\frac{1}{c},\frac{1}{d}$ are in A,P
Let $x$ be the first term and $y$ be the common difference of the A.P

$\Rightarrow \frac{1}{a}=x$; $\frac{1}{b}=x+y$; $\frac{1}{c}=x+2y$; $\frac{1}{d}=x+3y$
Now, $ab+bc+cd=\frac{1}{x(x+y)}+\frac{1}{(x+y)(x+2y)}+\frac{1}{(x+2y)(x+3y)}$
$\Rightarrow ab+bc+cd=\frac{(x+2y)(x+3y)+x(x+3y)+x(x+y)}{x(x+y)(x+2y)(x+3y)}=\frac{3(x^2+3xy+y^2)}{x(x+y)(x+2y)(x+3y)}$

$\Rightarrow ab+bc+cd=\frac{3(x+y)(x+2y)}{x(x+y)(x+2y)(x+3y)}=\frac{3}{x(x+3y)}$
Therefore, $ab+bc+cd=3ad$
Ans :B