Question

Let $a, a + r$ and $a + 2r$ be positive real numbers such that their product is $64$. Then the minimum value of $a + 2r$ is equal to

• A. 4
• B. 3
• C. 2
• D. 44563

Answer 1

Answer: (A)

4

Answer 2

We know $A M \geq G M$
$\frac{a+(a+r)+(a+2 r)}{3} \geq(a(a+r)(a+2 r))^{\frac{1}{3}}$
$\Rightarrow \frac{3(a+r)}{3} \geq(64)^{\frac{1}{3}}$
$\Rightarrow (a+r) \geq 4$
Also , $64=a(a+r)(a+2 r)$
$\Rightarrow 64 \geq(4-r) \times 4(r+4)$
$\Rightarrow 16 \geq 16-r^{2}$
$\Rightarrow r^{2} \leq 0$
$\therefore r=0$
Now, $a+2 r=4+0=4$