An over-ambitious AP

Algebra Level 4

Let ${a_{i}},0 \le i \le n$ be an arithmetic progression (AP) such that $a_{0} = 16$ and $a_{j} > 0$ for all $j \ge 0$ .

Also,every three consecutive terms of this AP satisfy the inequality $\displaystyle \dfrac{1}{a_{n}} \ge \dfrac{1}{\sqrt{a_{n+1} \cdot a_{n-1}}}.$

Find the value of $\displaystyle \sum_{j=0}^{99} \dfrac{\sqrt[4]{a_{j}}}{a_{j}}$ .

The answer is 12.5.

2 solutions

Rishabh Jain
Jun 10, 2016

$\dfrac{1}{a_{n}} \ge \dfrac{1}{\sqrt{a_{n+1} \cdot a_{n-1}}}$ $\implies \color{#D61F06}{a_n}\leq\sqrt{a_{n+1} \cdot a_{n-1}}$ $\implies \color{#D61F06}{\dfrac{a_{n+1}+a_{n-1}}2}\leq\sqrt{a_{n+1} \cdot a_{n-1}}$

$(\small{\color{#3D99F6}{\because a_{n-1},a_n,a_{n+1}\text{ form an AP}}})$

$\implies AM\leq GM~~(\small{of~a_{n-1},a_n,a_{n+1}})$

This is possible only when $AM=GM$ . Hence
$a_{n-1}=a_{n+1}$ which implies A.P is constant with each term as $16$ and summation is simply :-

$\dfrac{1}{(16)^{3/4}}\displaystyle \sum_{j=0}^{99}(1)=\dfrac{100}{8}=\boxed{12.5}$

Tamir Dror
Jun 25, 2016

I think I have an easier solution, will post it soon.

An over-ambitious AP

The answer is 12.5.

2 solutions

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