Question

If A be one A.M. and p, q be two G.M.'s between two numbers, then 2A is equal to

• A. $\frac{p^3+q^3}{pq}$
• B. $\frac{p^3-q^3}{pq}$
• C. $\frac{p^2+q^2}{2}$
• D. $\frac{pq}{2}$

Answer 1

The correct option is A

$\frac{p^3+q^3}{pq}$

(a) $\frac{p^3+q^3}{pq}$

Let the two positive numbers be a and b a, A and b are in A.P.

$\therefore$ 2A = a + b ....... (i)

Also, a, p, q and b are in G.P.

$\therefore r={\left(\frac{b}{a}\right)}^{\frac{1}{3}}$

Again, p = ar and a = $a{r}^2$. ...... (ii)

Now, 2A = a + b [From (i)]

$=a+a\left(\frac{b}{a}\right)$

$=a+a{\left({\left(\frac{b}{a}\right)}^{\frac{1}{3}}\right)}^3$

$=a+a{r}^3$

$=\frac{(ar{)}^2}{a{r}^2}+\frac{(a{r}^2{)}^2}{ar}$

$=\frac{p^2}{q}+\frac{q^2}{p}$ [Using (ii)]

$=\frac{p^3+q^3}{pq}$