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If the numbers $n−2,\ 4n−1$ and $5n+2$ are in A.P., find the value of $n$.
Given: The numbers $n-2,\ 4n-1$ and $5n+2$ are in A.P.
To do: To find the value of $n$.
Solution:
$n-2,\ 4n-1,\ 5n+2$ are in A.P.
First term, $a_1=n-2$
Second term, $a_2=4n-1$
Third term, $a_3=5n+2$
Common difference $d=a_2-a_1=a_3-a_2$
​
$\Rightarrow 4n-1-( n-2)=5n+2-( 4n-1)$
$\Rightarrow 4n-1-n+2=5n+2-4n+1$
$\Rightarrow 3n+1=n+3$
$\Rightarrow 2n=2$
$\Rightarrow n=\frac{2}{2}=1$
Thus, $n=1$.
Updated on: 10-Oct-2022
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