# 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?

**Solution:**

a_{n} = a + (n - 1)d is the nth term of AP.

Here, a is the first term, d is the common difference and n is the number of terms.

Incomes received by Subba Rao in the years 1995, 1996 and 1997 are 5000, 5200 and 5400 respectively.

This forms an AP with a = 5000 and d = 200

Let after n^{th} year, his salary be Rs 7000.

Hence, a_{n} = 7000

We know that the n^{th} term of an A.P. is a_{n} = a + (n - 1)d

By substituting above values, we get

7000 = 5000 + (n - 1) 200

200 (n - 1) = 7000 - 5000

n - 1 = 2000/200

n = 10 + 1

n = 11

Therefore, in the 11^{th} year, his income reached Rs 7000. This means after 10 years of 1995, that is, 1995 + 10 ⇒ 2005

Answer: In 2005, his income reached Rs 7000.

**Video Solution:**

## 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?

### Class 10 Maths NCERT Solutions Chapter 5 Exercise 5.2 Question 19 - Chapter 5 Exercise 5.2 Question 19:

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?

The year in which Subba Rao income reached Rs. 7000 is 2005, if Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year