Subba Rao started work in 1995 at an annual salary of ` 5000 and received an increment of ` 200 each year. In which year did his income reach ` 7000?
Step-by-Step Solution
Key Concept: The annual salaries form an arithmetic progression (AP) with first term $a = 5000$ and common difference $d = 200$. Use the $n^{\text{th}}$ term formula $a_n = a + (n-1)d$ to find the term that equals $7000$, then convert the term number to the corresponding year.
1. Identify the AP:
- First term (salary in 1995): $a = 5000$.
- Common difference (annual increment): $d = 200$.
2. Let $a_n$ be the salary in the $n^{\text{th}}$ year after 1995. The $n^{\text{th}}$ term of an AP is given by:
$$a_n = a + (n-1)d$$
3. Set $a_n = 7000$ and solve for $n$:
$$7000 = 5000 + (n-1)\times 200$$
$$7000 - 5000 = (n-1)\times 200$$
$$2000 = 200\,(n-1)$$
$$n-1 = \frac{2000}{200} = 10$$
$$n = 10 + 1 = 11$$
4. The first term corresponds to the year 1995, so the $n^{\text{th}}$ term corresponds to the year:
$$\text{Year} = 1995 + (n-1) = 1995 + 10 = 2005$$
5. Hence, Subba Rao's salary becomes `7000` in the year 2005.
Correct Answer: 2005