Question

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

Answer 1

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 \[{t_n} = a + \left( {n - 1} \right)d\]