
If $\dfrac{1}{{(p + q)}},{\text{ }}\dfrac{1}{{(r + p)}},{\text{ }}\dfrac{1}{{(q + r)}}$are in AP, then
$1)$p,q,r are in AP
$2)\;{{\text{p}}^2},{q^2},{r^2}$are in AP
$3)\;\dfrac{1}{p},\dfrac{1}{q},\dfrac{1}{r}$are in AP
$4)$None of these
Answer
493.2k+ views
Hint: In Arithmetic progression (AP) the difference between any two consecutive terms remains constant. Here we will take the difference of second and first term and will take difference between third and the second term and then will simplify for the required values.
Complete step-by-step answer:
Given that the three terms are in arithmetic progression, the difference between the two consecutive terms remains the same.
$\dfrac{1}{{(r + p)}} - \dfrac{1}{{(p + q)}} = \dfrac{1}{{(q + r)}} - \dfrac{1}{{(r + p)}}$
Make like terms together. When you move any term from one side to the opposite side then the sign of the term changes. Positive term becomes negative and vice versa.
$\dfrac{1}{{(r + p)}} + \dfrac{1}{{(r + p)}} = \dfrac{1}{{(q + r)}} + \dfrac{1}{{(p + q)}}$
When denominators are the same then numerators are combined. Also take LCM (least common multiple) on the right hand side of the equation.
$\dfrac{{1 + 1}}{{(r + p)}} = \dfrac{{(p + q) + (q + r)}}{{(q + r)(p + q)}}$
Simplify the above equation-
$\dfrac{2}{{(r + p)}} = \dfrac{{p + 2q + r}}{{(q + r)(p + q)}}$
Cross-multiply the above expression, where the numerator of one side is multiplied with the denominator of the opposite side and vice-versa.
$2(q + r)(p + q) = (r + p)(p + 2q + r)$
Simplify the above expression –
$2(pq + rp + {q^2} + rq) = (rp + 2rq + {r^2} + {p^2} + 2pq + rp)$
Multiply the term outside the bracket with the terms inside the bracket –
$2pq + 2rp + 2{q^2} + 2rq = 2rp + 2rq + {r^2} + {p^2} + 2pq$
Like terms with the same values and the same sign cancels each other.
$2{q^2} = {r^2} + {p^2}$
Hence, ${{\text{p}}^2},{q^2},{r^2}$are in Arithmetic progression.
Thus, from the given multiple choices – the second option is the correct answer.
So, the correct answer is “Option B”.
Note: Be clear with the difference between Arithmetic progression (AP) and geometric progression (GP). The difference between the terms in AP remains constant and the ratio between the terms remains the same in GP. Be careful about the sign convention, when there is a positive sign outside the bracket then the sign of the terms inside the bracket remains the same.
Complete step-by-step answer:
Given that the three terms are in arithmetic progression, the difference between the two consecutive terms remains the same.
$\dfrac{1}{{(r + p)}} - \dfrac{1}{{(p + q)}} = \dfrac{1}{{(q + r)}} - \dfrac{1}{{(r + p)}}$
Make like terms together. When you move any term from one side to the opposite side then the sign of the term changes. Positive term becomes negative and vice versa.
$\dfrac{1}{{(r + p)}} + \dfrac{1}{{(r + p)}} = \dfrac{1}{{(q + r)}} + \dfrac{1}{{(p + q)}}$
When denominators are the same then numerators are combined. Also take LCM (least common multiple) on the right hand side of the equation.
$\dfrac{{1 + 1}}{{(r + p)}} = \dfrac{{(p + q) + (q + r)}}{{(q + r)(p + q)}}$
Simplify the above equation-
$\dfrac{2}{{(r + p)}} = \dfrac{{p + 2q + r}}{{(q + r)(p + q)}}$
Cross-multiply the above expression, where the numerator of one side is multiplied with the denominator of the opposite side and vice-versa.
$2(q + r)(p + q) = (r + p)(p + 2q + r)$
Simplify the above expression –
$2(pq + rp + {q^2} + rq) = (rp + 2rq + {r^2} + {p^2} + 2pq + rp)$
Multiply the term outside the bracket with the terms inside the bracket –
$2pq + 2rp + 2{q^2} + 2rq = 2rp + 2rq + {r^2} + {p^2} + 2pq$
Like terms with the same values and the same sign cancels each other.
$2{q^2} = {r^2} + {p^2}$
Hence, ${{\text{p}}^2},{q^2},{r^2}$are in Arithmetic progression.
Thus, from the given multiple choices – the second option is the correct answer.
So, the correct answer is “Option B”.
Note: Be clear with the difference between Arithmetic progression (AP) and geometric progression (GP). The difference between the terms in AP remains constant and the ratio between the terms remains the same in GP. Be careful about the sign convention, when there is a positive sign outside the bracket then the sign of the terms inside the bracket remains the same.
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