If the numbers $n-2,\ 4n-1$ and $5n+2$ are in A.P., then find the value of $n$.

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

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

Solution:

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

$\because$ Common difference is the same in A.P.

Therefore, $( 4n-1)-( n-2)=( 5n+2)-( 4n-1)$   

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

$\Rightarrow 3n-3=n+3$

$\Rightarrow 3n-n=3+3$

$\Rightarrow 2n=6$

$\Rightarrow 2n=6$

$\Rightarrow n=\frac{6}{2}$

$\Rightarrow n=3$

Thus, $n=2$

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Updated on: 10-Oct-2022

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