 Welcome friends to another session on problem solving and we are taking up problems related to Sequence and series in this question. It's asked that if the eighth term of an AP is 31, okay And the 15th term is 16 more than 11th term So now they are giving you some kind of relationship between the terms, right? You have to find the AP What does finding the AP mean means the finding the AP means you first have to find out the first term and the common difference and Yes show some terms of that AP as well So this is what finding the AP means first term and common difference if you find out you can find out the AP, okay? Now so a sum of an AP is 31 So how to solve it so we will be using some knowledge of linear equations in this problem And hence whatever we have learned in the linear equation chapter linear equation in two variables would be applicable here, okay? So eighth term one of AP is 31. Okay, so first of all, let us say let us say That E1 is the first term, okay, and D is the common difference Okay, so this is what we are intending to find first term and the common difference now From the first line 8th term of an AP is 31. That means T8 is how much guys T8 is T1 plus 8 minus 1d. This is what we have been doing so far very easy T8 is 31 so 31 is equal to T1 We don't know what T1 is 8 minus 1 is 7 and this is D So this is the first equation we got even plus 70 Let us say this is equation number one or we can rearrange so that it becomes easier for us to solve later on So this implies T1 plus 7d is equal to 31 and let it be question number one Okay, now let's find out the another relationship So we have two variables T1 and D to be found out So if we get one more equation our job is done Now it's just 15th term is 16 more than the 11th term 15th term So let it be like this T15 Is this 16 more it is 16 more than the 11th term This is what they are saying so T11 plus 16 is T15 What is T15 guys T15 will be T1 plus 15 minus 1 times D and This is 16 plus T11 will be simply T1 plus 11 minus 1 right 11 minus 1d This is a new equation if you see this T1 and this T1 will disappear. So 14d Is equal to 16 plus 10d right now? So this implies 14 minus 10d if you rearrange you'll get D for 4d is 4d is 16 so D is 4 so Friends you have got one information that common difference is 4 So you can find out a now from any of the equations So let's take equation one from one you write from from one T1 will be how much 31 minus 7d correct 31 minus 7d 31 minus 7d is seven times 4 is 28. So this answer is 3 right So hence we got the we got the AP how now the AP would be so AP is nothing but first term Then second term is 33 plus 4 Third term will be 3 plus 2 times 4 4th term will be 3 plus 3 times 4 and so on and so forth. So hence this is 3 7 11 and 15 and so on this is the Given AP right. This is what they were asking for okay. So you understood what how to find out the AP so you just need to find out the first term and the common difference and you will get the AP and how to find First term and common difference through the given relationships in the question