Question

If a, b, c, d are in HP, then $ab+bc+cd$ is equal to
A. $3ad$
B. $(a+b)(c+d)$
C. $3ac$
D. none of these

Answer 1

Hint: Here we have given a, b, c, d are in HP. So we can use the general conditions of HP to obtain the values of a,b,c,d . If the reciprocals are in arithmetic progression, a series of values or numbers is called a harmonic progression. And after finding the values we can easily substitute in the equation to find the resultant value.

Formula used:
If a, b, c, d are in H.P then $\dfrac{1}{\mathrm{a}}, \dfrac{1}{\mathrm{~b}}, \dfrac{1}{\mathrm{c}}, \dfrac{1}{\mathrm{~d}}$ are in A,P

Complete Step by step solution:
If the reciprocal of the terms is in AP, a series of numbers is referred to as a harmonic progression. Simply said, if $\dfrac{1}{\mathrm{a}}, \dfrac{1}{\mathrm{~b}}, \dfrac{1}{\mathrm{c}}, \dfrac{1}{\mathrm{~d}}$ are in AP, then terms a, b, c, d are in HP.
Given, a, b, c, d are in H.P.
So $\dfrac{1}{\mathrm{a}}, \dfrac{1}{\mathrm{~b}}, \dfrac{1}{\mathrm{c}}, \dfrac{1}{\mathrm{~d}}$ are in A,P.

Let $x$ be the first term and $y$ be the common difference of the A.P
The standard deviation in the arithmetic progression is represented by $d$.
The distinction between a term's succeeding and preceding terms. When it comes to arithmetic progression, it is always constant or the same.
$\Rightarrow \dfrac{1}{a}=x ; \dfrac{1}{b}=x+y ; \dfrac{1}{c}=x+2 y ; \dfrac{1}{d}=x+3 y$
Now,
$a b+b c+c d=\dfrac{1}{x(x+y)}+\dfrac{1}{(x+y)(x+2 y)}+\dfrac{1}{(x+2 y)(x+3 y)}$
$\Rightarrow ab+bc+cd=\dfrac{(x+2y)(x+3y)+x(x+3y)+x(x+y)}{x(x+y)(x+2y)(x+3y)}$
$=\dfrac{3\left( {{x}^{2}}+3xy+{{y}^{2}} \right)}{x(x+y)(x+2y)(x+3y)}$
Simplifying,
$\Rightarrow ab+bc+cd=\dfrac{3(x+y)(x+2y)}{x(x+y)(x+2y)(x+3y)}$
$\Rightarrow a b+b c+c d=\dfrac{3(x+y)(x+2 y)}{x(x+y)(x+2 y)(x+3 y)}=\dfrac{3}{x(x+3 y)}$
Therefore, $a b+b c+c d=3 a d$
So the correct answer is option (A).

Note: Making the appropriate AP series is the first step in solving an issue involving harmonic progression. The nth term of an H.P. is given by $1/ [a + (n -1) d]$ in the same way that the nth term of an A.P is ${{a}_{n}}=a+(n-1)d$