 Now, here is another very interesting question says, the ratio of the sum of n terms of two APs is 7n plus 1 is to 4n plus 27. You have to find the ratio of their mth term. So what's given? Let's write there are two APs, first of all, two APs are there, two APs, okay, two APs are there, all right. And it's given, let's say a1, d1 and a2, d2 are first term and common difference of both APs, okay. So first assume this one, a1, d1 and a2, d2 are first terms and common difference of both APs, okay, right. Now it's given ratio of the sum of n terms, right. So let me write it as sn1 for the first AP and sn2 for second AP. And this is given as 7n plus 1 and 4n plus 27, right. First n terms of first AP is, you know, divide by first n terms of second AP, the sum of both these n terms is 7n plus 1 and 4n plus 27, okay. Now what is to be found out? We have to find out ratio of their mth terms. So we have to find out sm, sorry not s, am, am1, am for the first one and am for the second one is what is the question, okay. So now that we have summarized the question, let's try to find this out, okay. So let's first keep an eye on what is to be found out, am1 meaning a plus m minus 1, d1 for the first AP, second AP, a2 plus m minus 1, d2, right. This is the ratio of the mth term of both the APs. So this is what we have to achieve, okay. Now let's see what's given, sn1 is given, what is sn1? So sum of first n terms n by 2 twice a1 plus n minus 1, d1, correct. And the same thing for the second AP, n by 2 twice a2 plus n minus 1, d2, correct. And this is equal to 7n plus 1 divided by 4n plus 27. Now whenever such questions are there where you have to go from this format to this format, it's not a good idea to cross multiply and try and simplify because you see there are four variables a1, d1, a2, d2. So you will not be able to solve for all, that means you will not be able to find out a1, d1, a2, d2 because there is only one equation given. So hence you have to play with the format, that is whatever is required, let's see we can convert one format to the other format or not. Okay, so clearly this n by 2, n by 2 goes. So what's left? I can see I have 2 a1 but there is no 2 in this case, see no 2. So you have to somehow detach 2 from that a1. So what can I do? Can I not take 2 common and write like this a1 plus n minus 1 by 2, d1, okay and similarly in the denominator I can again extract 2 and write a2 plus n minus 1 by 2, d2, right and this is equal to 7n plus 1 by 4n plus 27, isn't it? Now if you see now our this step expression under LHS is looking somewhat similar to whatever we want, this 2 and 2 will go but there is a problem that there is n minus 1 by 2 somehow I want to get it converted to m, right? If that is equal to m then my problem is solved or m minus 1 rather. So can I not deliberately put it, what do I mean? Can I say let n minus 1 by 2 be equal to m minus 1, let, okay, when will that be possible? So this will be possible when n by 2 minus 1 by 2 is equal to m minus 1. That means how do I find n in terms of m? So what is n then? So n by n minus 1 will be equal to twice m minus 1 and hence n is equal to 1 plus 2m minus 2 that is 2m minus 1. So when n is 2m minus 1, right, this is possible, this thing is possible, right? So I am saying when n is equal to 2m minus 1 then n minus 1 by 2 will be equal to m minus 1 simply. So let us put n is equal to 2m minus 1 across the equation. So at n is equal to 2m minus 1, let this equation be called 1. At n is equal to 2m minus 1, 1 will become what? Simply put wherever you see n, put 2m minus 1. So you will get a1 plus 2m minus 1 minus 1 by 2d1, right? So let me just, yeah, so that it is in one frame and this is, second is a1, sorry, a2 plus 2m minus 1 minus 1 by 2d2 and this is equal to, now again here in terms of, in place of n, right, 2m minus 1 plus 1 and in the denominator, right, 4 into 2m minus 1 plus 1, right? What did we do? In this expression, if this is true for any n, I can substitute n for 1, 2, 3, 5, whatever and it will hold because this was given, this is given, isn't it? So if whatever value of n I put in the LHS, if I put the same thing in the RHS, it will hold. So I simply put n is equal to 2m minus 1 and what do I get? I get a1 plus, if you simplify this, you will get m minus 1d1 divided by a2 plus m minus 1d2 and here you will get 14m minus 7 plus 1 divided by 8m minus 4 plus 1, right? Minus 4, isn't it? So sorry, what was the, sorry, it is 27, it's not 4 into n plus 27, so you have to write 27, so there's a mistake here, this is 27, okay? So you have to write 27 here and simplify, what is this, 14m minus 6 divided by 8m minus or rather plus 23, right? It is 14m minus 6 divided by 8m plus 23 and what is there in the LHS? If you look closely, what is this? Isn't the numerator am for the first AP and the denominator is am for the second AP, right? And this is equal to 14m minus 6 by 8m plus 23 and this is exactly what you had to find out, ratio of mth term of both the APs, right? Hence, this is the, hence it is found out, right? This is what you have to find out, ratio of mth term of first AP and that of second AP, correct? So how did we solve this guys? So I told you, don't cross multiply the first equation which is the given constraint or a condition which you had, don't cross multiply and try finding a1d1, you will not be able to, why? Because there are four variables a1d1, a2d2. So hence that will not be helpful. Instead of that, I tried to transform one expression into the other by some manipulation on both sides and here the manipulation was, if n becomes 2m minus 1, then on the left hand side, you could get am that is mth term and accordingly change the ratio on the right hand side and you get the desired result.