JEE MainMathematicsSequences and Series
If the sum of the infinite series 3 + 5x + 7x^2 + 9x^3 + is 44 9 and |x| < 1 , then the value of x is :
Options
- A17 11
- B1 4
- C17 62
- D1 2
Correct answer
B. 1 4
Step-by-step solution
The given series is an arithmetico-geometric progression (AGP) with first term a = 3 , common difference d = 2 , and common ratio x . The sum of an infinite AGP for |x| S_ = a 1-x + dx (1-x)^2 Substituting the known values: 44 9 = 3 1-x + 2x (1-x)^2 44 9 = 3(1-x) + 2x (1-x)^2 44 9 = 3 - x 1 - 2x + x^2 Cross-multiplying gives: 44(1 - 2x + x^2) = 9(3 - x) 44 - 88x + 44x^2 = 27 - 9x 44x^2 - 79x + 17 = 0 Solving for x using the quadratic formula: x = 79 (-79)^2 - 4(44)(17) 2(44) x = 79 6241 - 2992 88 x = 79 3249 88 x =