Question

n-1P3:n+1P3 5:12 find n

Answer 1

The value of 'n' will be:  8

Explanation

Given equation is....

$^(^n^-^1^)P_{3} : ^(^n^+^1^)P_{3} = 5:12$

The formulas we will use are:  $^nP_{r}= \frac{n!}{(n-r)!}$  and $n! = n(n-1)(n-2)(n-3)...............$

First we will apply this formula and then solve.....

$^(^n^-^1^)P_{3} : ^(^n^+^1^)P_{3} = 5:12\\ \\ \frac{^(^n^-^1^)P_{3}}{^(^n^+^1^)P_{3}} = \frac{5}{12} \\ \\ \frac{\frac{(n-1)!}{(n-4)!}}{\frac{(n+1)!}{(n-2)!}}= \frac{5}{12}\\ \\ \frac{(n-1)!* (n-2)!}{(n-4)!*(n+1)!}=\frac{5}{12}\\ \\ \frac{[(n-1)!][(n-2)*(n-3)*(n-4)!]}{[(n-4)!][(n+1)*(n)*(n-1)!]}=\frac{5}{12}\\ \\ \frac{(n-2)(n-3)}{n(n+1)}=\frac{5}{12}\\ \\ \frac{n^2-5n+6}{n^2+n}=\frac{5}{12}\\ \\ 12n^2-60n+72=5n^2+5n\\ \\ 12n^2-60n+72-5n^2-5n=0\\ \\ 7n^2-65n+72=0\\ \\ 7n^2-56n-9n+72=0$

$7n(n-8)-9(n-8)=0\\ \\ (n-8)(7n-9)=0$

Using zero-product property, we will get......

$n-8=0\\ n=8\\ \\ and\\ \\ 7n-9=0\\ 7n=9\\ n=\frac{9}{7}$

($n=\frac{9}{7}$ is ignored as (n-1) and (n+1) must be greater than 3)

Answer 2

Answer:

Step-by-step explanation:

Mark my ans as brainliest plz

I have done by taking as 1/9.

So u will put 5/7 in place of that and solve it plzzz