Question

Which of the following can be terms (not necessarily consecutive) of any A.P. ?

• A. $1,6,19$
• B. $\sqrt{2},\sqrt{50},\sqrt{98}$
• C. $\log 2,\log 16,\log 128$
• D. $\sqrt{2},\sqrt{3},\sqrt{7}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $1,6,19$
B $\sqrt{2},\sqrt{50},\sqrt{98}$
C $\log 2,\log 16,\log 128$

Let $a,b,c$ be the $p^{th},q^{th},r^{th}$ terms of an A.P. respectively whose first term is $A$ and the common difference is $D$.
Then $a=A+(p-1)D,b=A+(q-1)D,c=A+(r-1)D$
$\Rightarrow c-b=(r-q)D$ and $b-a=(q-p)D$
$\therefore \frac{c-b}{b-a}=\frac{r-q}{q-p}\left(\text{a rational}\right)$
This means $\frac{c-b}{b-a}$ must be a rational number.

For $1,6,19$
$\frac{c-b}{b-a}=\frac{19-6}{6-1}$
which is a rational number.

For $\sqrt{2},\sqrt{50},\sqrt{98}$
$\frac{c-b}{b-a}=\frac{\sqrt{98}-\sqrt{50}}{\sqrt{50}-\sqrt{2}}=\frac{7\sqrt{2}-5\sqrt{2}}{5\sqrt{2}-\sqrt{2}}=\frac{1}{2}$
which is a rational number.

For $\log 2,\log 16,\log 128$
$\frac{c-b}{b-a}=\frac{\log 128-\log 16}{\log 16-\log 2}=\frac{7\log 2-4\log 2}{4\log 2-\log 2}=1$
which is a rational number.

But for $\sqrt{2},\sqrt{3},\sqrt{7}$
$\frac{c-b}{b-a}=\frac{\sqrt{7}-\sqrt{3}}{\sqrt{3}-\sqrt{2}}$
which is not a rational number.