Question

# Find the value of 11th term of an A.P.: 4,9,14,...

Given series: $4,9,14,...$
Difference between the first and the second term is $5$, and the second and third term is also $5$.
Therefore, common difference in this series is $5$,
${t_n} = 4 + \left( {n - 1} \right)5 = 5n - 1$
For $n = 11,{t_n} = 5\left( {11} \right) - 1 = 54$