 Hello and welcome to the session. In this session we discussed the following question which says for what value of p are 2p-1, 7 and 3p, 3 consecutive terms of an ap. We know that a given list of numbers say a1, a2, a3 and so on is an ap if ak plus 1 minus ak is same for different values of k that is a2 minus a1 is equal to a3 minus a2 is equal to 1 so on. This is the key idea to be used in this question. Let's proceed with the solution now. We are given the three terms of an ap 2p minus 1, 7 and 3p these three terms are in ap and we have to find the value for p. Now if these three terms are in ap so using the key idea we have a2 minus a1 would be equal to a3 minus a2 this means 7 minus 2p minus 1 the whole would be equal to 3p minus 7 further 7 minus 2p plus 1 is equal to 3p minus 7 this gives us 8 minus 2p is equal to 3p minus 7 that is 3p plus 2p is equal to 8 plus 7 further we get 5p is equal to 15 now dividing both sides by 5 we get 5p upon 5 is equal to 15 upon 5 5 5 gets cancelled and 5 3 times is 15 so this gives us p is equal to 3 so we get the value for p as 3 for which the given three terms are in ap so this is the final answer this completes the session hope you have understood the solution of this question.