Question

The continued product of three numbers in G.P is 216 and the sum of the product of them in pairs is 156; then the number of prime numbers less than the highest of the three numbers is

Answer 1

Let the three numbers in G.P is $\frac{a}{r},a,ar$
It is given, $\frac{a}{r}.a.ar=216$
$\Rightarrow a^3=216$ $\therefore a=6$
Again, $\frac{a}{r}.a+a.ar+ar.\frac{a}{r}=156$
$\Rightarrow a^2(\frac{1}{r}+r+1)=156$
$\Rightarrow \frac{1+r+r^2}{r}=\frac{156}{36}=\frac{13}{3}$
$\Rightarrow 3+3r+3r^2=13r$ $\Rightarrow 3r^2-10r+3=0$
$\Rightarrow (3r-1)(r-3)=0$ $\therefore r=\frac{1}{3}or3$
If $r=\frac{1}{3},$ then the numbers are $18,6,2$ and if $r=3$, then the numbers are $2,6,18$

The highest among the three numbers is $18$

Prime numbers less than $18$ are $2,3,5,7,11,13,and17$

Hence, the number of prime numbers less than the highest of the three numbers is $7$.