 Hi and welcome to the session. I am Deepika here. Let's discuss the question. This term of the AP 121, 117, 113 and so on is its first negative term. Hint, fine and for A n is less than 0. So, let's start the solution. We know that and that term of an AP that is A n is equal to A plus n minus 1 d where A is the first term, d is the common difference. Let us use this key idea for the above question. Now, let's start the solution. Given AP is 121, 117, 113 and so on. Here, A is equal to 121 which is the first term and d is equal to the common difference 117 minus 121 which is equal to minus 4. Let of the AP is its first negative term. Then, n is less than 0. This implies A plus n minus 1 d is less than 0. This implies 121 plus n minus 1 into d is minus 4 is less than 0. This implies 121 minus 4n plus 4 is less than 0. This implies 125 minus 4n is less than 0 or 125 is less than 4n or greater than 125. This implies n is greater than 125 by 4 or n is greater than 31 and a 1 by 4. Now, this is our natural number. So, we will take n is greater than or equal to 32 because n is our natural number. The answer for the above question is that 32nd term of an AP is its first negative term. We can check our solution also. We know that let us check is equal to A plus n minus 1 d that is 30 d which is equal to A was our 121 plus 30 into minus 4 which is equal to 121 minus 120 which is equal to 1. So, this is positive. Now, when we will take A32 that is equal to A plus 31 d this is equal to 121 plus 31 into minus 4. This is equal to 121 minus 124 which is equal to minus 3. So, A31 is positive and after 31 we have taken 32 so A32 is negative. So, our answer for the above question is that 32nd term of the AP is its first negative term. I hope the question is clear to you. Bye and have a nice day.