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# If $\dfrac{1}{{q + r}}$, $\dfrac{1}{{r + p}}$, and $\dfrac{1}{{p + q}}$ are in AP, then which of the following is true?(a). p, q, r are in AP(b). ${p^2},{q^2},{r^2}$ are in AP(c). $\dfrac{1}{p},\dfrac{1}{q},\dfrac{1}{r}$ are in AP(d). None of these  Hint: In an AP the consecutive terms differ by the same constant. The average of the first and third number is the second number. Hence, use this to equate and then find the relation.

Complete step-by-step solution -
An arithmetic progression (AP) is a sequence of numbers whose consecutive terms differ by a constant number. This constant number is called the common difference.
If three numbers are in AP, then the average of the first and the third number gives the second number.
We are given that $\dfrac{1}{{q + r}}$, $\dfrac{1}{{r + p}}$, and $\dfrac{1}{{p + q}}$ are in AP. Hence, the average of $\dfrac{1}{{q + r}}$ and $\dfrac{1}{{p + q}}$ gives $\dfrac{1}{{r + p}}$. Hence, we have as follows:
$\dfrac{1}{2}\left( {\dfrac{1}{{q + r}} + \dfrac{1}{{p + q}}} \right) = \dfrac{1}{{r + p}}$
Taking 2 to the right-hand side of the equation, we have:
$\dfrac{1}{{q + r}} + \dfrac{1}{{p + q}} = \dfrac{2}{{r + p}}$
Simplifying, we have:
$\dfrac{{p + q + q + r}}{{(q + r)(p + q)}} = \dfrac{2}{{r + p}}$
Evaluating the denominator, we have:
$\dfrac{{p + 2q + r}}{{{q^2} + rp + rq + pq}} = \dfrac{2}{{r + p}}$
Cross multiplying, we have:
$(p + 2q + r)(r + p) = 2({q^2} + rp + rq + pq)$
Simplifying the above expression, we have:
${p^2} + 2qp + rp + pr + 2qr + {r^2} = 2{q^2} + 2rp + 2rq + 2pq$
Simplifying further, we have:
${p^2} + 2qp + 2rp + 2qr + {r^2} = 2{q^2} + 2rp + 2rq + 2pq$
Canceling the common terms, we have:
${p^2} + {r^2} = 2{q^2}$
Dividing both sides by 2, we have:
$\dfrac{{{p^2} + {r^2}}}{2} = {q^2}$
Hence, ${q^2}$ is the average of ${p^2}$ and ${r^2}$.
Hence, ${p^2}$, ${q^2}$, and ${r^2}$ are in AP.
Hence, the correct answer is option (b).

Note: You can also find the common difference of the given AP and use the fact that adding a constant number to all the terms of the AP still gives an AP to check which of the following options is an AP. Here we can use that the difference between two consecutive terms equates to the difference of next two consecutive terms like if a,b,c are in A.P then b-a = c-b.

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