 Hello friends, welcome again to another problem-solving session on sequence and series and we are Taking up problems related to middle terms of an arithmetic progression This question says that show that in an AP with odd number of terms, right odd number of terms Twice the middle term is equal to the sum of first and last terms, okay? So this is what we have to prove that there is an AP Which is having odd number of terms you have to prove that twice that middle term middle term of that AP two times The middle term of that in that AP will be equal to the first and the last terms Let's see how to go about it. So we will start with this. Let us say Let us say That the number of terms that the number of Terms in the given AP in The given AP is equal to N, right and N is odd. So hence can we say N is equal to 2k plus one for sum K With some K, which is a non-negative Integer non-negative Integer, okay? So that's what we are saying. So N is equal to 2k plus one, right? So we know that in this case the middle term middle term, how do we find out middle term middle term is simply for odd number of terms, we know that this is T and Plus one by two. So hence it is T 2k plus one plus one by two, which is equal to T K plus one, right? E K plus one is the middle term, okay? So E K plus one is how much by our formula of the nth term We can know that this is T one the first term plus K plus one minus one times D Isn't it for D to be the common difference of this AP? So we can say E K Plus one is equal to two, no, sorry not two, E one Plus K times D Okay Isn't it? Now, so twice the middle term in this case will be this, right? Which is nothing but two T one plus two KD Right, so keep this in mind. So twice the middle term is two T one plus two KD now first term We have to add first term and the last term Right, so first term and the last term is T one plus last term is T one plus n minus one D Correct, isn't it? So that means this is T one two T one plus N minus one D, right? And we know that n is two K plus one two T one plus n What was n guys? We had assumed n to be equal to two K plus one, right? So I can write two K plus one here two K plus one then minus one D this minus one is because of this one Okay, so hence what do I get two T one plus two KD? See they are same. So hence this can be written as this is two times T K plus one, which is nothing but two times The middle term Right, this is how we could prove that in an AP with odd number of terms Twice the middle term is equal to the sum of the first and the last terms