Question

If a, b, c are in A.P. p, q, r are in H.P and ap, bq, cr are in G.P then:
A. $$\dfrac{p}{r} +\dfrac{r}{p} =\dfrac{c}{a} +\dfrac{a}{c}$$
B. $$\dfrac{p}{q} +\dfrac{q}{p} =\dfrac{c}{a} +\dfrac{a}{c}$$
C. $$\dfrac{p}{r} +\dfrac{r}{p} =\dfrac{b}{c} +\dfrac{c}{b}$$
D. none of these

Answer 1

Hint: In this question it is given that a, b, c are in A.P. p, q, r are in H.P and ap, bq, cr are in G.P then we have to find the relation between the elements. So to find the solution we need to know that if a, b, c are in A.P then we can write,
$$b=\dfrac{a+c}{2}$$, where b is the arithmetic mean.
If p, q, r are in H.P then $$\dfrac{2}{q} =\dfrac{1}{p} +\dfrac{1}{r}$$,
Where q is the harmonic mean.
And if x, y, z are in G.P then, $$y^{2}=xz$$ where y is the geometric mean.

Complete step-by-step answer:
Here given a, b, c are in A.P.
So by the above formula we can write,
$$b=\dfrac{a+c}{2}$$.............(1)
Also p, q, r are in H.P, so by the above formula we can write,
$$\dfrac{2}{q} =\dfrac{1}{p} +\dfrac{1}{r}$$
$$\Rightarrow \dfrac{2}{q} =\dfrac{r+p}{pr}$$
$$\Rightarrow \dfrac{q}{2} =\dfrac{pr}{r+p}$$
$$\Rightarrow q=\dfrac{2pr}{r+p}$$.............(2)
Now since ap, bq, cr are in G.P then by the formula of geometric mean we can write,(where x=ap, y=bq, z=cr)
$$\left( bq\right)^{2} =\left( ap\right) \left( rc\right) $$
$$\Rightarrow b^{2}q^{2}=acpr$$
Now by putting the value of $$b^{2}$$ and $$q^{2}$$ in the above equation, we get,
$$\Rightarrow \left( \dfrac{a+c}{2} \right)^{2} \left( \dfrac{2pr}{r+p} \right)^{2} =acpr$$
$$\Rightarrow \dfrac{\left( a+c\right)^{2} }{2^{2}} \cdot \dfrac{\left( 2rp\right)^{2} }{\left( r+p\right)^{2} } =acpr$$
$$\Rightarrow \dfrac{\left( a+c\right)^{2} }{4} \cdot \dfrac{4r^{2}p^{2}}{\left( r+p\right)^{2} } =acpr$$
$$\Rightarrow \dfrac{\left( a+c\right)^{2} }{\left( r+p\right)^{2} } \cdot p^{2}r^{2}=acpr$$
$$\Rightarrow \dfrac{\left( a+c\right)^{2} }{\left( r+p\right)^{2} } =\dfrac{acpr}{p^{2}r^{2}}$$ [dividing both sides by $$p^{2}r^{2}$$]
$$\Rightarrow \dfrac{\left( a+c\right)^{2} }{\left( r+p\right)^{2} } =\dfrac{ac}{pr}$$
$$\Rightarrow \dfrac{\left( a+c\right)^{2} }{ac} =\dfrac{\left( r+p\right)^{2} }{pr}$$
Now as we know that $$\left( x+y\right)^{2} =x^{2}+2xy+y^{2}$$,
So by using this identity we can write the above equation as,
$$\dfrac{a^{2}+2ac+c^{2}}{ac} =\dfrac{r^{2}+2rp+p^{2}}{pr}$$
$$\Rightarrow \dfrac{a^{2}}{ac} +\dfrac{2ac}{ac} +\dfrac{c^{2}}{ac} =\dfrac{r^{2}}{pr} +\dfrac{2rp}{pr} +\dfrac{p^{2}}{pr}$$ [ by separating term wise]
$$\Rightarrow \dfrac{a}{c} +2+\dfrac{c}{a} =\dfrac{r}{p} +2+\dfrac{p}{r}$$
$$\Rightarrow \dfrac{a}{c} +\dfrac{c}{a} +2=\dfrac{r}{p} +\dfrac{p}{r} +2$$
$$\Rightarrow \dfrac{a}{c} +\dfrac{c}{a} =\dfrac{r}{p} +\dfrac{p}{r} +2-2$$
$$\Rightarrow \dfrac{a}{c} +\dfrac{c}{a} =\dfrac{r}{p} +\dfrac{p}{r}$$
$$\Rightarrow \dfrac{p}{r} +\dfrac{r}{p} =\dfrac{c}{a} +\dfrac{a}{c}$$
Hence the correct option is option A.

Note: While solving this type of question you need to know that for any series, either it is A.P, G.P or H.P, the middle term of any consecutive three terms of a series is called the mean value, so for an A.P this mean is called arithmetic mean and for G.P it is geometric mean and for H.P it is Harmonic mean.