 In this question, it's given that if mth term of an AP is 1 upon n and the nth term is 1 upon m show that sum of mn terms, m times n, it's actually m times n, so right is half mn plus 1, so what does this mean? So it's like mth term like 6th term, let's say m is equal to 6, so 6th term of an AP is let's say 1 upon 7 and 7th term is let's say 1 upon 6, that is what mn and just an example to make it clear to you. So 6th term is 1 upon 7 and 7th term is 1 upon 6, then you have to show that mns term that is 6 into 7 that is 40 second term, 40 second term is given by half into 6 into 7 plus 1 something like that. This is just an example, I don't know whether m is 6 or n is 7, but just to explain what is mnth term, mth term and nth term right, so if you consider 6 as m and 7 as n, so 6th term is 1 upon n that is 1 upon 7 and 7th term is 1 upon 6, so mnth term will be 6 into 7 that is 40 second term, that's an just an example to show you or make you understand this concept or question question. Now how to go about it, so direct application of formula it requires I believe, so mth term as an AP is 1 upon i n, so you know how to find out mth term, so 1 upon n is first term, let's say first term is a plus what m minus minus 1 d where a is equal to first term and d is common difference right, so this is given, now 1 upon m similarly will be a plus n minus 1 d, this is 1 and this is 2, now we can eliminate either of the variables, so let's say we do 1 minus 2, so 1 minus 2 will give you 1 minus 1 by n minus 1 upon m is equal to a and a will get cancelled and it will be m minus 1 d minus n minus 1 d which is clearly m minus n d correct, now the left hand side take the common denominator mn, so it will become like that this is equal to m minus n d, I hope how do you calculate common denominator, they should not be an issue, so mn will be the common denominator the denominator, so now where there is 1 on top of n you can have m and minus n right, this is how you have to do this, so mn divided by this n will leave you m that m multiplied by this 1 will give you this one here, similarly mn divided by this m is n and n multiplied by this 1 comes here and becomes n right, this is what we have done, I hope you know this just for those who you know at times struggle in finding the common denominator and hence equivalent fraction, so hence now assuming m minus n is not equal to 0, why is this needed, because if m is equal to n then you can't cancel these two terms, so assuming assuming, so let assuming m is not equal to n which is fair enough assumption because we are talking about ms term being 1 by n and ns term being 1 by m and all that, so assuming that this is not true, so hence what will you get, you will get d is equal to 1 upon mn, assuming that what, m is not equal to n right, d is 1 upon mn, so one variable you could figure out, let's find out a now, so what is a from 1 if you see from first equation, a is 1 upon n minus m minus 1d from 1, okay so this is 1 upon n minus m minus 1 and d is 1 by mn okay, so this becomes 1 by n minus 1 by n again because m times 1 by mn will give you 1 by n and plus 1 by mn, so this again goes so a is 1 upon mn right, so we got a as 1 upon mn and is equal to d both are same, so what to find out, we have to find out mnf term and sum of mnf term right, so smn we have to find out, what will this be, it will be simply mn by 2 for number of terms by 2 multiplied by twice a, what was a 1 upon mn minus sorry plus mn minus 1 times d 1 by mn, okay what is this formula guys, we are assuming applying this n upon 2 twice a plus n minus 1d, so wherever n is there I have just deployed mn, okay so you'll get that and a is 1 upon mn and d is 1 upon mn, so now solve this, so this becomes mn and if you multiply the first this thing it is 2 by mn correct and then plus mn by 2 and within brackets what is left inside, the second term is mn minus 1 by mn correct, this is simply 1 plus if you see you will get half 1 plus half mn minus 1 correct which is equal to 1 plus mn by 2 minus half which is equal to mn by 2 plus 1 by 2 which is equal to half mn plus 1 right hence proved hence proved right, this is what you have to do in this question