Question

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

Answer 1

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 \Rightarrow8n - 2 = n - 2 + 5n + 2$$ By addition of $$n - 2$$ and $$5n + 2$$ we get $\Rightarrow8n - 2 = 6n Rewrite the equation after simplification we get $\Rightarrow$$$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$$