Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?

Subba Rao’s starting salary = Rs 5000
It means, first term = a = 5000

He gets an increment of Rs 200 after every year.

Therefore, common difference = d = 200

His salary after 1 year = 5000 + 200 = Rs 5200
His salary after two years = 5200 + 200 = Rs 5400

Therefore, it is an AP of the form 5000, 5200, 5400, 5600… , 7000

We want to know in which year his income reaches Rs 7000.

Using formula \(a_n = a + (n - 1)d\) , to find \(n^{th}\) term of arithmetic progression,

\(\Rightarrow \) 7000 = 5000 + (n - 1) (200)
\(\Rightarrow \) 7000 = 5000 + 200n - 200
\(\Rightarrow \) 7000 – 5000 + 200 = 200n
\(\Rightarrow \) 2200 = 200n
\(\Rightarrow \) n = 11

It means after 11 years, Subba Rao’s income would be Rs 7000.