If the numbers $n-2$, $4n-1$, and $5n+2$ are in A.P. , Find the value $n$.

Given: Numbers $n-2$, $4n-1$, and $5n+2$ are in A.P.

To do: To find the value of $n$.

Solution: 

$\because n-2,\ 4n-1,\ and\ 5n+2$  are in A.P.

$\therefore$ Common difference $d=( 4n-1)-( n-2)=( 5n+2)-(4n-1)$

$\Rightarrow 4n-1-n+2=5n+2-4n+1$

$\Rightarrow 3n+1=n+3$

$\Rightarrow 3n-n=3-1$

$\Rightarrow 2n=2$

$\Rightarrow n=\frac{2n}{2}$

$\Rightarrow n=1$

Thus, $n=1$. 

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Updated on: 10-Oct-2022

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