Question

If $x=∑_{n=0}^∞a^n,y=∑_{n=0}^∞b^n,z=∑_{n=0}^∞c^n$, where a, b, c are in A.P. and |a| < 1, |b| < 1, |c| < 1, abc≠ 0, then :

• A. x, y, zare in A.P.
• B. x, y, zare in G.P.
• C. $\frac{1}{x},\frac{1}{y},\frac{1}{z}$ are in A.P
• D. $\frac{1}{x}+\frac{1}{y}+\frac{1}{z} = 1-(a+b+c)$

Answer 1

Answer: (C)

$\frac{1}{x},\frac{1}{y},\frac{1}{z}$ are in A.P

Answer 2

If $x=∑_{n=0}^∞a^n,y=∑_{n=0}^∞b^n,z=∑_{n=0}^∞c^n$= $\frac{1}{1-c}$

Now,

$a, b, c→ AP$

$1 – a, 1 – b, 1 – c→ AP$

$\frac{1}{1−a},\frac{1}{1−b},\frac{1}{1−c}→HP$

$x, y, z→ HP$

$⇒ \frac{1}{x},\frac{1}{y},\frac{1}{z}$ are in $A.P$

Hence, the correct option is (C) : $\frac{1}{x},\frac{1}{y},\frac{1}{z}$ are in A.P