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?

Question

Homi · Accepted Answer

We will use the formula for the nth term of an Arithmetic Progression to find the number of years.

Step 1 — List the given information

Let's write down what we know from the problem. The starting salary is the first term. The annual increment is the common difference. The target salary is the $n$ -th term.

$\text{First term } (a) = \text{₹ } 5000$

$\text{Common difference } (d) = \text{₹ } 200$

$n\text{-th term } (a_n) = \text{₹ } 7000$

Diagram 1

Step 2 — Find the number of years (n)

We use the formula for the $n$ -th term of an AP. The formula is $a_n = a + (n-1)d$ . Let's substitute the values we know into the formula.

$7000 = 5000 + (n-1)200$

Now, let's solve this equation for $n$ . First, subtract 5000 from both sides.

$7000 - 5000 = (n-1)200$

$2000 = (n-1)200$

Next, divide both sides by 200.

$\frac{2000}{200} = n-1$

$10 = n-1$

Finally, add 1 to both sides to find $n$ .

$n = 10 + 1$

$\boxed{n = 11}$

Step 3 — Determine the year

The value $n=11$ means it is the 11th year. The first salary was received in 1995. So, the 11th salary will be received in the year $1995 + (11-1)$ .

$\text{Year} = 1995 + (n-1)$

$\text{Year} = 1995 + (11-1)$

$\text{Year} = 1995 + 10$

$\boxed{\text{Year} = 2005}$

Answer

His income reached ₹ 7000 in the year 2005.

Question 19

Answer

More questions in Exercise 5.2