 Hi and welcome to the session. Let us discuss the following question. The question says find four numbers forming the geometric progression in which the third term is greater than the fourth term by nine and the second term is greater than the fourth by eighty. Let's now begin with the solution. Let A be the first term, R be the common ratio of a GP. Now in the question it is given that third term is greater than the first term by nine. So according to this condition third term that is T3 is equal to first term that is T1 plus 9. Now the second condition given to us is second term is greater than the fourth by 18. So according to this condition T2 that is second term is equal to T4 that is fourth term plus 18. 3 is equal to T1 plus 9 implies AR squared is equal to A plus 9 right and T2 is equal to T4 plus 18 implies AR is equal to ARQ plus AT. Let's name this equation as equation number one and this as two. Now on multiplying we get ARQ is equal to AR plus 9R. Let's name this equation as equation number three. Now from 2 and 3 we have AR minus 18 is equal to AR plus 9R and this implies 9R is equal to minus 18 and this implies R is equal to minus 2. Now we will find the value of A. So let's now substitute the value of R in 1. On substituting the value of R in 1 we get A into minus 2 whole square is equal to A plus 9 and this implies 4A is equal to A plus 9 and this implies 3A is equal to 9 and this implies A is equal to 3. Now the first term of the GP that is T1 is equal to A which is equal to 3. Second term that is T2 is equal to AR and this is equal to 3 into minus 2. This is equal to minus 6 and the third term that is T3 is equal to AR square and this is equal to 3 into minus 2 square and this is equal to 12 and the fourth term that is T4 is equal to ARQ and this is equal to 3 into minus 2 whole q and this is equal to minus 24. Hence the required four numbers forming a Gp are 3 minus 6, 12 and minus 24. This is our required answer. So this completes the fashion. Bye and take care.