 Hi and welcome to the session. My name is Rishi and I am going to help you to solve the following question. The question is, Subharao 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? Let us start the solution now. It is given that Subharao's initial salary is equal to Rs. 5000 and each year's increment is equal to Rs. 200. So, we can write Subharao's initial salary is equal to Rs. 5000 and increment of each year is equal to Rs. 200. So, the required AP is 5000, 5200, 5400 as he is getting the increment of Rs. 200. So, the common difference in the AP is equal to Rs. 200. Now, in this AP, A is equal to 5000, A is the first term of AP and common difference V is equal to 200. The salary of Subharao will be Rs. 7000 in the end of the year. So, we can write Subharao's salary will be Rs. 7000 in the end of the year. Now, we know the electric term of the AP is given by A n equal to A plus n minus 1 multiplied by D where A is the first term of AP and D is the common difference. So, we can write A n is equal to 5000 plus n minus 1 multiplied by D where D is equal to 200. We know the salary of Subharao would be Rs. 7000 in the end of the year. So, we can write A n is equal to 7000. Now, substituting for A n, 7000 we get 7000 is equal to 5000 plus n minus 1 multiplied by 200. Now, this implies 7000 minus 5000 is equal to 200 multiplied by n minus 1. This implies 2000 is equal to 200 multiplied by n minus 1. Now, dividing both sides by 200 we get 10 is equal to n minus 1. Or we can write n minus 1 is equal to 10. This implies n is equal to 10 plus 1 equal to 11. So, we get n is equal to 11. So, we can write his income will reach to be 7000 in the 11th year. So, our required answer is 11th year. This completes the session. Hope you understood the session. Have a nice day.