Question

Let a,b and c be in AP and $|a|<1,|b|<1|c|<1$.
If $x=1+a+a^2+......\infty$
$y=1+b+b^2+......\infty$
$z=1+c+c^2+......\infty$ then x, y and z are in.

• A. AP
• B. GP
• C. HP
• D. None of these

Answer 1

The correct option is C HP

$x=1+a+a^2+.....\infty =\frac{1}{1-a}$

$y=1+b+b^2+.....\infty =\frac{1}{1-b}$

and $z=1+c+c^2+.....\infty =\frac{1}{1-c}$

Since, a,b and c are in AP.

$\Rightarrow 1-a,1-band1-c$ are also in AP.

$\Rightarrow \frac{1}{1-a},\frac{1}{1-b}and\frac{1}{1-c}$ are in HP.

$\therefore$ x, y and z are in HP.