Sequences and Series 5 Question 20
11. Let $A _n=\frac{3}{4}-\frac{3}{4}^{2}+\frac{3}{4}^{3}+\ldots+(-1)^{n-1} \frac{3}{4}^{n}$,
$B _n=1-A _n$. 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 $n$.
$$ \therefore \quad l+n=2 m $$
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.
$\therefore \quad G _1=l r, G _2=l r^{2}, G _3=l r^{3}, n-l r^{4} \Rightarrow r=\frac{n}{l}^{\frac{1}{4}}$
Now, $G _1^{4}+2 G _2^{4}+G _3^{4}=(l r)^{4}+2\left(l r^{2}\right)^{4}+\left(l r^{3}\right)^{4}$
$$ \begin{aligned} & =l^{4} \times r^{4}\left(1+2 r^{4}+r^{6}\right)=l^{4} \times r^{4}\left(r^{4}+1\right)^{2} \\ & =l^{4} \times \frac{n}{l} \frac{n+l}{l}{ }^{2}=l n \times 4 m^{2}=4 l m^{2} n \end{aligned} $$
