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?
11th year
Step 1: Recognise the AP.
Subba Rao’s salary each year increases by a fixed amount (Rs 200), so his annual salaries form an arithmetic progression (AP).
First term (salary in the first year 1995):
\[a = 5000\]
Common difference (yearly increment):
\[d = 200\]
We want to know in which year his salary becomes Rs 7000. Let that be the \(n\)th year of his service.
Step 2: Use the nth term formula of an AP.
For an AP, the nth term is:
\[a_n = a + (n - 1)d\]
Here, we want \(a_n = 7000\). So:
\[7000 = 5000 + (n - 1) \cdot 200\]
Step 3: Solve the equation for \(n\).
Subtract 5000 from both sides:
\[7000 - 5000 = (n - 1) \cdot 200\]
\[2000 = 200(n - 1)\]
Divide both sides by 200:
\[n - 1 = \dfrac{2000}{200} = 10\]
So,
\[n = 10 + 1 = 11\]
Step 4: Interpret the result.
The salary reaches Rs 7000 in the 11th year of service.
If he started in 1995 (1st year), then the 11th year is:
\[1995 + 10 = 2005\]
Conclusion: His income reaches Rs 7000 in the 11th year (i.e., in the year 2005).