Question

If the product of $3$ consecutive numbers in G.P. is $216$ and the sum of their products in pairs is $156$, then the smallest term in the three is

• A. $2$
• B. $6$
• C. $\frac{1}{4}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $2$

Let the three numbers be $\frac{a}{r},a,ar$
Now, the product
$a^3=216\Rightarrow a=6$
Sum of the products in pairs
$\frac{a}{r}\times a+a\times ar+\frac{a}{r}\times ar=156\Rightarrow 36(r^2+r+1)=156r\Rightarrow 3r^2-10r+3=0\Rightarrow (3r-1)(r-3)=0\Rightarrow r=\frac{1}{3},3$

Hence, the three terms are $2,6,18$ or $18,6,2.$