COMEDK · Maths · 5. Sequences and Series
If S is the sum to infinity of a decreasing geometric progression with common ratio \(x\) such that \(|x| <1 ; x \neq 0\). The ratio of fourth term to the second term is \(\dfrac{1}{16}\) and the ratio of the third term to the square of the second term is \(\dfrac{1}{9}\). Then the value of S is
- A 12
- B 36
- C 7.2
- D 48
Answer & Solution
Correct Answer
(A) 12
Step-by-step Solution
Detailed explanation
Let the first term of the geometric progression be \(a\) and the common ratio be \(x\). The terms are \(a, ax, ax^2, ax^3, \dots\). Given the ratio of the fourth term to the second term is \(\dfrac{1}{16}\), we have \(\dfrac{ax^3}{ax} = x^2 = \dfrac{1}{16}\). Since the…
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