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Question: If *a*, *b*, *c* are in A.P. and \|*a*\|, \|*b*\|, \|*c*\| \< 1, and \(x = 1 + a + a ^ { 2 } + \ldot...

If a, b, c are in A.P. and |a|, |b|, |c| < 1, and x=1+a+a2+x = 1 + a + a ^ { 2 } + \ldots \infty, y=1+b+b2+y = 1 + b + b ^ { 2 } + \ldots \infty , z=1+c+c2+z = 1 + c + c ^ { 2 } + \ldots \ldots \infty

Then x, y, z shall be in

A

A.P.

B

G.P.

C

H.P.

D

None of these

Answer

H.P.

Explanation

Solution

x=1+a+a2+=11ax = 1 + a + a ^ { 2 } + \ldots \infty = \frac { 1 } { 1 - a }

z=1+c+c2+.=11cz = 1 + c + c ^ { 2 } + \ldots . \infty = \frac { 1 } { 1 - c }

Now, a, b, c are in A.P.

⇒ 1 – a, 1 – b, 1 – c are in A.P.

11a,11b,11c\frac { 1 } { 1 - a } , \frac { 1 } { 1 - b } , \frac { 1 } { 1 - c } are in H.P.

Therefore x, y, z are in H.P.