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

The Correct Option is C

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