Verified Solution • 2025–26 Session moderate

QJEEM22Ma985b: JEE Main 2022 Mathematics - Class 11 JEEM22Ma985b Solution: , where a, b, c are in A.P. and

x=∑n=0∞an,y=∑n=0∞bn,z=∑n=0∞cnx = \sum\limits_{n = 0}^\infty {{a^n},y = \sum\limits_{n = 0}^\infty {{b^n},z = \sum\limits_{n = 0}^\infty {{c^n}} } } , where a, b, c are in A.P. and |a| < 1, |b| < 1, |c| < 1, abc ≠\ne 0, then :

Multiple Choice Options

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

Step-by-Step Verified Solution

x=∑n=0∞an=11−a; y=∑n=0∞bn=11−b; z=∑n=0∞cn=11−cx = \sum\limits_{n = 0}^\infty {{a^n} = {1 \over {1 - a}};\,y = \sum\limits_{n = 0}^\infty {{b^n} = {1 \over {1 - b}};\,z = \sum\limits_{n = 0}^\infty {{c^n} = {1 \over {1 - c}}} } }

Now,

a, b, c →\to AP

1 −- a, 1 −- b, 1 −- c →\to AP

11−a, 11−b, 11−c→HP{1 \over {1 - a}},\,{1 \over {1 - b}},\,{1 \over {1 - c}} \to HP

x, y, z →\to HP

∴\therefore 1x,1y,1z→AP{1 \over x},{1 \over y},{1 \over z} \to AP

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring