 Hello and welcome to this session. Let us understand the following question today. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively. Now let us write the solution. Given to us is, second term is equal to 14, third term is equal to 18 and n is equal to 51 and we have to find the sum of first 51 terms. Now, from second and third term we can find the value of A and D as A2 is equal to A plus D where A2 is 14 is equal to A plus D, let us name it as Equation 1. Similarly, A3 is equal to A plus 2D which gives A3 is 18 is equal to A plus 2D, this is our equation 2. Now solving equation 1 and 2 we get A plus D is equal to 14 and A plus 2D is equal to 18. Now on subtraction, do not forget to change the signs while subtracting. So, this and this gets cancelled, D minus 2D will give us minus D which is equal to 14 minus 18 will give us minus 14 minus 4. So, this and this gets cancelled. So, it implies D is equal to 4. Now, substituting the value of D is equal to 4 in equation 1 we get 14 is equal to A plus 4 which implies A is equal to 14 minus 4 which implies A is equal to 10. Now we have to find Sn that is S51 and we know Sn is equal to n by 2 multiplied by 2A plus n minus 1D. Now we have to find the value of S51 which is equal to n is 51 by 2 multiplied by 2 into 10 plus 51 minus 1 multiplied by D which is equal to 4. Now which is equal to 51 by 2 multiplied by 20 plus 50 into 4 which is equal to 51 by 2 20 plus 200 which is equal to 51 by 2 multiplied by 220. Now, this gets cancelled by 110. So, it is equal to 5610. Therefore, S51 is equal to 5610. Hence, S51 is equal to 5610 is our required answer. I hope you understood the question. Bye and have a nice day.