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]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
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]

Ans.
(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, $α = 5$

Want Us to Email you About Special Offers & Updates?