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Question: If \[\dfrac{1}{{\left( {{\text{q}} + {\text{r}}} \right)}}\], \[\dfrac{1}{{\left( {{\text{r}} + {\te...

If 1(q+r)\dfrac{1}{{\left( {{\text{q}} + {\text{r}}} \right)}}, 1(r+p)\dfrac{1}{{\left( {{\text{r}} + {\text{p}}} \right)}} and 1(p+q)\dfrac{1}{{\left( {{\text{p}} + {\text{q}}} \right)}} are in A.P, then
A. p, q and r are in A.P
B. p2{{\text{p}}^2}, q2{{\text{q}}^2} and r2{{\text{r}}^2} are in A.P
C. 1p\dfrac{1}{{\text{p}}}, 1q\dfrac{1}{{\text{q}}} and 1r\dfrac{1}{{\text{r}}} are in A.P
D. None of the above

Explanation

Solution

Hint: Let us use the condition that the common difference of A.P is always the same and twice of the second term is equal to the sum of the first and third term of the A.P.
As we know that a, b and c are the three terms that are in A.P. Then the common difference of this A.P will be b – a and c – b.

Complete step-by-step answer:
And as we know that the common difference of the A.P is always equal. So, b – a = c – b.
So, applying this condition to the given A.P. We get,
1(r+p)\dfrac{1}{{\left( {{\text{r}} + {\text{p}}} \right)}} – 1(q+r)\dfrac{1}{{\left( {{\text{q}} + {\text{r}}} \right)}} = 1(p+q)\dfrac{1}{{\left( {{\text{p}} + {\text{q}}} \right)}} – 1(r+p)\dfrac{1}{{\left( {{\text{r}} + {\text{p}}} \right)}} (1)
So, now we have to solve the above equation.
So, taking LCM in LHS and RHS of the above equation. We get,
q+r−r−p(r+p)(q+r)\dfrac{{{\text{q}} + {\text{r}} - {\text{r}} - {\text{p}}}}{{\left( {{\text{r}} + {\text{p}}} \right)\left( {{\text{q}} + {\text{r}}} \right)}} = r + p−p−q(p+q)(r+p)\dfrac{{{\text{r + p}} - {\text{p}} - {\text{q}}}}{{\left( {{\text{p}} + {\text{q}}} \right)\left( {{\text{r}} + {\text{p}}} \right)}}
q−p(r+p)(q+r)\dfrac{{{\text{q}} - {\text{p}}}}{{\left( {{\text{r}} + {\text{p}}} \right)\left( {{\text{q}} + {\text{r}}} \right)}} = r−q(p+q)(r+p)\dfrac{{{\text{r}} - {\text{q}}}}{{\left( {{\text{p}} + {\text{q}}} \right)\left( {{\text{r}} + {\text{p}}} \right)}}
Now multiplying both sides of the above equation by (r+p)\left( {{\text{r}} + {\text{p}}} \right). We get,
q−p(q+r)\dfrac{{{\text{q}} - {\text{p}}}}{{\left( {{\text{q}} + {\text{r}}} \right)}} = r−q(p+q)\dfrac{{{\text{r}} - {\text{q}}}}{{\left( {{\text{p}} + {\text{q}}} \right)}}
So, now for solving the above equation. We cross multiplying the above equation.
(q−p)(p+q)\left( {{\text{q}} - {\text{p}}} \right)\left( {{\text{p}} + {\text{q}}} \right) = (r−q)(q+r)\left( {{\text{r}} - {\text{q}}} \right)\left( {{\text{q}} + {\text{r}}} \right)
Now solving the above equation. We get,
qp – p2{{\text{p}}^2} + q2{{\text{q}}^2} – pq = rq – q2{{\text{q}}^2} + r2{{\text{r}}^2} – qr
q2{{\text{q}}^2} – p2{{\text{p}}^2} = r2{{\text{r}}^2} – q2{{\text{q}}^2}
Now adding q2{{\text{q}}^2} + p2{{\text{p}}^2} to both sides of the above equation. We get,
2q2{{\text{q}}^2} = p2{{\text{p}}^2} + r2{{\text{r}}^2}
Now as we know that if three terms are in A.P the twice of second term is written as sum of first and third term.
So, p2{{\text{p}}^2}, q2{{\text{q}}^2} and r2{{\text{r}}^2} are in A.P.
Hence, the correct option will be B.

Note: Whenever we come up with this type of problem we have to form the equation by using the condition of A.P that the common difference of A.P is the difference if two consecutive terms and common difference always remain the same. And after solving this equation we will get another A.P if possible. This will be the easiest and efficient way to find the solution of the problem.