 Hello and welcome to another problem-solving session on sum of n terms first n terms of an arithmetic progression the given questions is find the sum of 20 terms of an AP in which third term is seven and Seventh term is two more than twice of its third term so we'll have to Frame some equations first write the relationships and then try install this problem So what's given sum of we have to find out? Let's find out you know solution is Let's find out. What is the objective solution for this particular solution objective of this problem is Find the sum of first 20 terms. So we have to find out s 20 Right s 20. What will s 20 be s 20 will be equal to 20 by 2 within brackets twice the first term plus 20 minus 1 times d Right, so we have to basically find out a and e which will be found out by the given conditions our third term is 7 a 3 is equal to 7 this implies a 3 what is a 3 guys a plus 2d isn't it and my a plus n minus 1d is a n This is 7 so we get first relationship Right, but this is one equation two variables. We can't solve for a and d You know again, we cannot get one solution exact solution. So let's go let's go for the second condition seventh term Right is two more than so is is given by equal to two more than so two more than so plus thrice of its third term thrice third term a 3 correct, this is the second equation This is the second equation. So let's try and solve Right now a 3 is anyways given a 7. I believe yes a 3 is given So clearly a 7 is 2 plus 3 into 7 So 23 so a 7 is 23 and what is a 7 by the way a 7 is a plus 6d is equal to 23 This is equation number 2 Okay, so can we do 2? Minus 1 let's subtract the second equation first equation from the second one You will get 4d A and a will get cancelled 6d minus 2d is 4d. So this is the second equation This is the first equation if you wish you can write it here so that it becomes easier for you to view So a plus 4d is 7 and I subtracted both of them and I'm writing the result here So 4d sorry a plus how much it was 2d not 4 2d so if you subtract you'll get 4d is equal to 23 minus 7 is 16 Right, so clearly D comes out to be 16 upon 4 4 Right D is known now. We have to just find out a so from one we can find out a from one a is equal to 7 minus 2d right and this is 7 minus 4 into 2 So minus 1 so we got a as well So now a and D both are known so sum of 20 terms will be known We can deploy the values here this equation and we are done. So Let me write here itself. So hence SN or s20 in this case s20 s20 is simply 20 by 2 is 10 into 2 times a so 2 times minus 1 Plus 20 minus 1 is 19 times D, which is 4 Isn't it so What this what will this be? So this is 10 into Minus 2 plus 76 Right, so 74 into 10 7 4 0 This is the solution so some of first 20 terms is 720 740 sorry, so what was the learning? How did we solve this problem once again? Just recap We were asked to find out some of 20 terms so some of 20 terms We know the formula 20 by 2 twice a plus n minus 1d But we didn't know a and D so we use the other two given conditions to find a and D and The conditions were given in terms of nth term, right? So we use the formula nth term is a plus n minus 1d Solve the two equations found out a and D deploy it back into the first equation here and we got s20 s7