Question:
Let \(A_n=\left(\frac{3}{4}\right)-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^3-\ldots+(-1)^{n-1}\left(\frac{3}{4}\right)^n\) and \(B_n=1-A_n\).
Then, the least odd natural number p, so that \(B_n>A_n\), for all \(n\ge p\), is :
Let \(A_n=\left(\frac{3}{4}\right)-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^3-\ldots+(-1)^{n-1}\left(\frac{3}{4}\right)^n\) and \(B_n=1-A_n\).
Then, the least odd natural number p, so that \(B_n>A_n\), for all \(n\ge p\), is :
Then, the least odd natural number p, so that \(B_n>A_n\), for all \(n\ge p\), is :
Show Hint
\(A_n\) is a GP with ratio \(-\frac{3}{4}\); \(B_n>A_n\) means \(A_n<\frac{1}{2}\). Check even and odd n separately.
Updated On: Nov 11, 2026
- 9
- 7
- 11
- 5
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The Correct Option is B
Solution and Explanation
\(A_n\) is a GP with first term 3/4 and ratio -3/4: \(A_n=\frac{\frac{3}{4}\left(1-(-\frac{3}{4})^n\right)}{1+\frac{3}{4}}=\frac{3}{7}\left(1-(-\tfrac{3}{4})^n\right)\).
\(B_n>A_n\iff A_n<\frac{1}{2}\iff 1-(-\tfrac{3}{4})^n<\frac{7}{6}\iff(-\tfrac{3}{4})^n>-\frac{1}{6}\).
For even n the left side is positive, so this always holds.
For odd n it becomes \((\tfrac{3}{4})^n<\frac{1}{6}\). \((\tfrac{3}{4})^5=\frac{243}{1024}\approx0.237>\frac{1}{6}\) fails, \((\tfrac{3}{4})^7=\frac{2187}{16384}\approx0.133<\frac{1}{6}\) holds, and larger odd n also hold.
So the condition holds for every n except n = 1, 3, 5. The least odd p with the condition true for all \(n\ge p\) is 7.
Answer: (2) 7
\(B_n>A_n\iff A_n<\frac{1}{2}\iff 1-(-\tfrac{3}{4})^n<\frac{7}{6}\iff(-\tfrac{3}{4})^n>-\frac{1}{6}\).
For even n the left side is positive, so this always holds.
For odd n it becomes \((\tfrac{3}{4})^n<\frac{1}{6}\). \((\tfrac{3}{4})^5=\frac{243}{1024}\approx0.237>\frac{1}{6}\) fails, \((\tfrac{3}{4})^7=\frac{2187}{16384}\approx0.133<\frac{1}{6}\) holds, and larger odd n also hold.
So the condition holds for every n except n = 1, 3, 5. The least odd p with the condition true for all \(n\ge p\) is 7.
Answer: (2) 7
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