Question

If the expression $\dfrac{1}{q+r},\dfrac{1}{r+p},\dfrac{1}{p+q}$ are in A.P., then:
A. p, q, r are in A.P.
B. ${{p}^{2}},{{q}^{2}},{{r}^{2}}$ are in A.P.
C. $\dfrac{1}{p},\dfrac{1}{q},\dfrac{1}{r}$ are in A.P.
D. None of these

Answer 1

- Hint:If three numbers (a, b, c) are in Arithmetic Progression then the common difference between terms is equal. It means b – a and c – b are equal to each other. Now, get the relation in p, q, r to get the answer.

Complete step-by-step solution -

It is given that $\dfrac{1}{q+r},\dfrac{1}{r+p},\dfrac{1}{p+q}$ are in A.P. so, we need to determine the relation in p, q and r.
As we know that differences in successive terms are equal in A.P. which is known as the common difference of an A.P.
Hence, if three numbers a, b, c are in A.P. then
b – a = c – b……………..(i)
So, terms $\dfrac{1}{q+r},\dfrac{1}{r+p},\dfrac{1}{p+q}$ are given in A.P.
Hence from equation (i), we get
$\dfrac{1}{r+p}-\dfrac{1}{q+r}=\dfrac{1}{p+q}-\dfrac{1}{r+p}$
Now, taking L.C.M in both the sides, we get
$\begin{aligned} & \dfrac{\left( q+r \right)-\left( r+p \right)}{\left( r+p \right)\left( q+r \right)}=\dfrac{\left( r+p \right)-\left( p+q \right)}{\left( p+q \right)\left( r+p \right)} \\\ & \Rightarrow \dfrac{q+r-r-p}{\left( r+p \right)\left( q+r \right)}=\dfrac{r+p-p-q}{\left( p+q \right)\left( r+p \right)} \\\ & \Rightarrow \dfrac{q-p}{\left( r+p \right)\left( q+r \right)}=\dfrac{r-q}{\left( p+q \right)\left( r+p \right)} \\\ \end{aligned}$
Cancelling (r + p) from both the sides, we get
$\dfrac{q-p}{q+r}=\dfrac{r-q}{p+q}$
On cross multiplying, we get
$\begin{aligned} & \left( q-p \right)\left( p+q \right)=\left( r-q \right)\left( q+r \right) \\\ & \Rightarrow \left( q-p \right)\left( q+p \right)=\left( r-q \right)\left( r+q \right)..........\left( ii \right) \\\ \end{aligned}$
Now, we can use an algebraic identity given as
$\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$
Hence, equation (ii) can be written as
${{q}^{2}}-{{p}^{2}}={{r}^{2}}-{{q}^{2}}.....(iii)$
Now, adding ${{p}^{2}}+{{q}^{2}}$ to both the sides, we get
$\begin{aligned} & {{q}^{2}}-{{p}^{2}}+{{p}^{2}}+{{q}^{2}}={{r}^{2}}-{{q}^{2}}+{{p}^{2}}+{{q}^{2}} \\\ & 2{{q}^{2}}={{r}^{2}}+{{p}^{2}}...............\left( iv \right) \\\ \end{aligned}$
Now, from the equation (ii), and (iii), we can observe that ${{p}^{2}},{{q}^{2}},{{r}^{2}}$ are in A.P, as it is satisfying condition of equation (i).
Hence, option (B) is the correct answer.

Note: Another approach to get relations among p, q, r would be using relation $2b=a+c$, if a, b, c are in A.P. As we know if a, b, c are in A.P. then $b-a=c-b\Rightarrow 2b=a+c$. So, one can apply $2b=a+c$ relation to get the required result.
Don’t miss any term p, q and r while doing calculation. And do not change the positions of the given terms in the problem while using the relation, $b-a=c-b\Rightarrow 2b=a+c$ otherwise the whole solution will ruin.