If P n - 1 : P n = 11 : 21 , 2 n - 1 2 n + 1 then n 2 + n + 15 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 ${}^{2 n + 1}{P_{n - 1} : {}^{2 n - 1}{P_{n} = 11 : 21,}}$ then $n^{2} + n + 15$ is equal to ______ . [2023]

Ans.
(45)

Given, ${}^{2 n + 1}{P_{n - 1}} :^{2 n - 1} P_{n} = 11 : 21$

$\Rightarrow \frac{(2 n + 1)!}{(2 n + 1 - n + 1)!} \times \frac{(n - 1)!}{(2 n - 1)!} = \frac{11}{21}$

$\Rightarrow \frac{(2 n + 1)!}{(n + 2)!} \times \frac{(n - 1)!}{(2 n - 1)!} = \frac{11}{21}$

$\Rightarrow \frac{(2 n + 1) (2 n) (2 n - 1)!}{(n + 2) (n + 1) (n) (n - 1)!} \times \frac{(n - 1)!}{(2 n - 1)!} = \frac{11}{21}$

$\Rightarrow \frac{(2 n + 1) (2 n)}{(n + 2) (n + 1) (n)} = \frac{11}{21} \Rightarrow \frac{4 n + 2}{n^{2} + 3 n + 2} = \frac{11}{21}$

$\Rightarrow 84 n + 42 = 11 n^{2} + 33 n + 22 \Rightarrow 11 n^{2} - 51 n - 20 = 0$

$\Rightarrow (n - 5) (11 n + 4) = 0 \Rightarrow n = 5, \frac{- 4}{11}$

The value of $n$ can never be negative.

So, $n = 5$

$∴$ $n^{2} + n + 15 = {(5)}^{2} + 5 + 15 = 45$

Want Us to Email you About Special Offers & Updates?