Question

The number of ways in which three distinct numbers can be chosen from the set $S=\{10,11,12,\dots ,100\}$ 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 $a,ar,a{r}^2$.
we have $a\ge 10$ and $a{r}^2\le 100$
$\Rightarrow r^2\le 10$
$\Rightarrow r=2,3(\because r>1$ and also an integer $)$
For $r=2,a{r}^2\le 100\Rightarrow a\le \frac{100}{4}=25$
Hence, $10\le a\le 25$
$\therefore$ For $r=2$, a can take 16 values.
$\{10,20,40;11,22,44\dots ...25,50,100\}$
Now for $r=3,a\le \frac{100}{9}\Rightarrow a\le 11$
Also $10\le a\le 11$
$\{10,30,90;11,33,99\}$
$\therefore$ For $r=3$, a can take 2 values.
Hence total number of ways $=18$