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# Find $n,$ if $n - 2,$$4n - 1$ and $5n + 2$ are in the A.P

Last updated date: 22nd Jun 2024
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Hint: Firstly know about the arithmetic progression. Then we use the concept of the arithmetic progression.After that we calculate the value of the $n$. Then substitute the value of the $n$ in the $n - 2,4n - 1$ and $5n + 2$.

Formula used: If three numbers $a,b$ and $c$ are in A.P. then
$2b = a + c$

Complete step-by-step solution:
It is given that $n - 2,4n - 1$ and $5n + 2$ are in A.P. then we use the concept of arithmetic progression
According to the concept
$\Rightarrow$$2\left( {4n - 1} \right) = n - 2 + 5n + 2$
$4n - 1$ is multiplied by $2$ we get
$\Rightarrow$$8n - 2 = n - 2 + 5n + 2$
By addition of $n - 2$ and $5n + 2$ we get
$\Rightarrow$$8n - 2 = 6n$
Rewrite the equation after simplification we get
$\Rightarrow$$8n - 6n - 2 = 0$
Substract $6n$ from $8n$ we get
$\Rightarrow$$2n - 2 = 0$
Rewrite the equation after simplification we get $2n = 2$
$2$ is divided by $2$we get
$\Rightarrow$$\dfrac{2}{2} = 1$
Hence the value of $n$ is $1$
Substitute the value of $n$ in $n - 2,4n - 1$ and $5n + 2$ we get
$\ n - 1 \\ 1 - 1 = 0 \$
Value of $4n - 1$ is
$4n - 1 = 4 \times 1 - 1 = 3$
Value of $5n + 2$ is
$5n + 2 = 5 \times 1 + 2 = 7$

Hence the value of $n$ is 1 and numbers are $0,3$ and $7$

Note: Arithmetic progression is a sequence whose terms increase or decrease by a fixed number. Fixed number is called the common difference.
If $a$ is the first term and $d$ is the common difference , then arithmetic progression can be written as $a,a + d,a + 2d................a + \left( {n - 1} \right)d$
${n^{th}}$ term of the arithmetic progression ${t_n} = a + \left( {n - 1} \right)d$