Algebra

Progression & Series

Medium

The product of 3 integers A, B, and C is 4096. The geometric mean of B and C is 32, whilst that of A and B is 8. Calculate C.

✅ Correct Option: 1

The geometric mean of two numbers $x$ and $y$ is $\sqrt{x \cdot y}$ . So if the GM is known, the product of the two numbers is $GM^2$ .

GM of $B$ and $C$ is $32$ :

$BC = 32^2 = 1024$

GM of $A$ and $B$ is $8$ :

$AB = 8^2 = 64$

Since $ABC = 4096$ :

$A \times (BC) = 4096$

$A \times 1024 = 4096$

$A = 4$

From $AB = 64$ :

$4 \times B = 64$

$B = 16$

From $BC = 1024$ :

$16 \times C = 1024$

$C = 64$

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