If a is the greatest term in the sequence a n = n 3 n 4 + 147 , n = 1 , 2 , 3 , … , then is equal to ________. [2023]

Question

If $a_{α}$ is the greatest term in the sequence $a_{n} = \frac{n^{3}}{n^{4} + 147}, n = 1, 2, 3, \dots,$ then $α$ is equal to ________. [2023]

Smart Achievers · Accepted Answer

(5)

We have, $a_{n} = \frac{n^{3}}{n^{4} + 147}$

Differentiate both sides w.r.t. $n$ , we get

$a_{n}' = \frac{(n^{4} + 147) 3 n^{2} - n^{3} \cdot 4 n^{3}}{{(n^{4} + 147)}^{2}}$

Put $a_{n}^{'} = 0$

$\Rightarrow \frac{n^{2} {3 (n^{4} + 147) - 4 n^{4}}}{{(n^{4} + 147)}^{2}} = 0$

$\Rightarrow 3 n^{4} + 441 - 4 n^{4} = 0 \Rightarrow n^{4} = 441 \Rightarrow n^{2} = 21 \Rightarrow n = \sqrt{21}$

$\Rightarrow 4 < \sqrt{21} < 5 \Rightarrow a_{5} > a_{4}$

So, $α = 5$

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