# 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?

**Solution:**

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

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

Since, there's an increment of ₹ 200 every year, the 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 ₹ 7000.

Hence, aₙ = 7000

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

By substituting the values in the above equation, 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 ₹ 7000. This means after 10 years of 1995, that is, 1995 + 10 ⇒ 2005

In 2005, his income reached ₹ 7000.

**☛ Check: **NCERT Solutions for Class 10 Maths Chapter 5

**Video Solution:**

## 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?

Class 10 Maths NCERT Solutions Chapter 5 Exercise 5.2 Question 19

**Summary:**

Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In the year 2005, his income reaches ₹ 7000.

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