Question

The number of ways in which three distinct numbers can be chosen from the set $S =\{10,11,12, \ldots, 100\}$ such that they form a G.P. with common ratio an integer greater than 1 , is

• A. 16
• B. 18
• C. 19
• D. 20

Answer 1

Answer: (B)

Let common ratio of G.P. be $r$ and first term is a. 3 number are $a , ar , ar ^2$.
we have $a \geq 10$ and $ar ^2 \leq 100$
$\Rightarrow r ^2 \leq 10$
$\Rightarrow r=2,3(\because r>1$ and also an integer $)$
For $r=2, a r^2 \leq 100 \Rightarrow a \leq \frac{100}{4}=25$
Hence, $10 \leq a \leq 25$
$\therefore$ For $r =2$, a can take 16 values.
$\{10,20,40 ; 11,22,44 \ldots . . .25,50,100\}$
Now for $r =3, a \leq \frac{100}{9} \Rightarrow a \leq 11$
Also $10 \leq a \leq 11$
$\{10,30,90 ; 11,33,99\}$
$\therefore$ For $r =3$, a can take 2 values.
Hence total number of ways $=18$