Question 6

This problem involves a sequence where a fixed amount is added each year.

Step 1 — Identify the given values

Let's find the starting salary. The initial salary is ₹5,00,000. This is our first term, $a$.

Let's find the annual increase. The annual increment is ₹20,000. This is our common difference, $d$.

We want to find when the salary reaches ₹7,00,000. This is the $n$-th term, $t_n$.

Step 2 — Apply the AP formula

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

$₹7,00,000 = ₹5,00,000 + (n - 1)₹20,000$

Now, we solve for $n$.

$₹7,00,000 - ₹5,00,000 = (n - 1)₹20,000$

$₹2,00,000 = (n - 1)₹20,000$

Let's divide both sides by ₹20,000.

$\frac{₹2,00,000}{₹20,000} = n - 1$

$10 = n - 1$

Now, we add 1 to both sides.

$n = 10 + 1$

$\boxed{n = 11}$

Step 3 — Interpret the result

The value $n=11$ means that the salary of ₹7,00,000 is reached in the 11th year. This means it took 10 increments to reach this salary. Each increment happens after one year of work. So, 10 years of work passed.

Answer

Harish's income reached ₹7,00,000 after 10 years.