 Hi and welcome to the session. My name is Shashi and I am going to help you to solve the following question. Question is, let me remind the AP whose third term is 16 and seventh term exceeds the fifth term by 12. First of all, let us understand the key idea to solve the given question. The nth term of an AP with first term A and common difference D is given by A n is equal to A plus n minus 1 multiplied by D where A is the first term of AP and D is the common difference of AP. Now, let us start with the solution. Now, to find the AP you will have to find the first term of the AP A and then find the common difference or we need to find A and D that is the first term and the common difference to find the AP. Now, we are given in the question that third term is 16. Now, we know nth term is given by as we have already read in key idea, nth term of the AP is equal to A plus n minus 1 multiplied by D. Now, substituting 3 for n we get third term is equal to A plus 3 minus 1 multiplied by D. This implies third term is equal to A plus 2 D. Now, we are given that third term is equal to 16. So, we can write 16 is equal to A plus 2 D as we are given third term is equal to 16. Similarly, we can find the fifth term of the AP that is equal to A plus 5 minus 1 multiplied by D where A is the first term of the AP and D is the common difference. So, we can write fifth term of the AP is equal to A plus 4 D. Similarly, we can find the seventh term of the AP that is equal to A plus 7 minus 1 multiplied by D where A is the first term of AP and D is the common difference. So, we get seventh term of the AP is equal to A plus 16. Now, we are given in the question that seventh term exceeds the fifth term by 12. So, we can write according to the question seventh term of the AP minus fifth term of the AP must be equal to 12. Now, you know seventh term is equal to A plus 60 as we have obtained here and fifth term is equal to A plus 4 D. So, for fifth term we substitute A plus 4 D. Now, this implies A plus 60 minus A minus 4 D is equal to 12. This implies 2 D is equal to 12. A and A will get subtracted and 60 minus 4 D is equal to 2 D. So, we get 2 D is equal to 12. Now, this implies D is equal to 12 upon 2 which is equal to 6. So, therefore we get D is equal to 6. Now, we know A plus 2 D is equal to 16. So, let us name this equation as 1. Now, we will substitute the value of D is equal to 6 in the equation 1 to get the value of A that is the first term of the AP. Now, substituting D is equal to 6 in equation 1 we get A plus 2 multiplied by 6 is equal to 16. This implies A plus 12 is equal to 16. This implies A is equal to 16 minus 12 is equal to 4. So, we get A is equal to 4. So, required AP can be written as 4, 4 plus 6 is equal to 10, 10 plus 6 is equal to 16, 16 plus 6 is equal to 22. Or simply we can write it as 4, 10, 16, 22. So, our final answer is 4, 10, 16, 22 is the required AP. This completes the session. Hope you understood the session. Take care and goodbye.