SATHEE: Sequences and Series 5 Question 20

Sequences and Series 5 Question 20

11. Let $A_{n} = \frac{3}{4} - {\frac{3}{4}}^{2} + {\frac{3}{4}}^{3} + \dots + (- 1)^{n - 1} {\frac{3}{4}}^{n}$ ,

$B_{n} = 1 - A_{n}$ . Find a least odd natural number $n_{0}$ , so that $B_{n} > A_{n}, \forall n \geq n_{0}$ .

(2006, 6M)

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Solution:

Given, $m$ is the AM of $l$ and $n$ .

$∴ l + n = 2 m$

and $G_{1}, G_{2}, G_{3}$ are geometric means between $l$ and $n$ .

$l, G_{1}, G_{2}, G_{3}, n$ are in GP.

Let $r$ be the common ratio of this GP.

$∴ G_{1} = l r, G_{2} = l r^{2}, G_{3} = l r^{3}, n - l r^{4} \Rightarrow r = {\frac{n}{l}}^{\frac{1}{4}}$

Now, $G_{1}^{4} + 2 G_{2}^{4} + G_{3}^{4} = (l r)^{4} + 2 {(l r^{2})}^{4} + {(l r^{3})}^{4}$

$\begin{aligned} = l^{4} \times r^{4} (1 + 2 r^{4} + r^{6}) = l^{4} \times r^{4} {(r^{4} + 1)}^{2} \\ = l^{4} \times \frac{n}{l} \frac{n + l}{l}^{2} = l n \times 4 m^{2} = 4 l m^{2} n \end{aligned}$

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Sequences and Series 5 Question 20

11. Let An=34−342+343+…+(−1)n−134n,

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11. Let $A_{n} = \frac{3}{4} - {\frac{3}{4}}^{2} + {\frac{3}{4}}^{3} + \dots + (- 1)^{n - 1} {\frac{3}{4}}^{n}$ ,