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JEE Mains · Maths · STD 11 - 8. sequence and series
Let \({A_n} = \left( {\frac{3}{4}} \right) - {\left( {\frac{3}{4}} \right)^2} + {\left( {\frac{3}{4}} \right)^3} - ..... + {\left( { - 1} \right)^{n - 1}}{\left( {\frac{3}{4}} \right)^n}\) and \(B_n \,= 1 - A_n\) . Then, the least odd natural number \(p\) , so that \({B_n} > {A_n}\), for all \(n \geq p\) is
- A \(5\)
- B \(7\)
- C \(11\)
- D \(9\)
Answer & Solution
Correct Answer
(B) \(7\)
Step-by-step Solution
Detailed explanation
\({A_n} = \left( {\frac{3}{4}} \right) - {\left( {\frac{3}{4}} \right)^2} + {\left( {\frac{3}{4}} \right)^3} - ...... + {\left( { - 1} \right)^{n - 1}}{\left( {\frac{3}{4}} \right)^n}\) Which is a \(G.P.\) with \(a = \frac{3}{4}'r = \frac{{ - 3}}{4}\) and number of terms \(=n\)…
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