 Hi and welcome to the session. I'm Kanigan and I'm going to help you to solve the following question. The question says which of all the following sequences c part is 1 by 3, 1 by 9, 1 by 27 and so on is 1 by 1 9 6 8 3. Let's now begin with the solution. Given geometric progression is 1 by 3, 1 by 9, 1 by 27 and so on. Let of the given GP 1 by 1 9 6 8 3. Now in the given GP the first term that is A is equal to 1 by 3 and R is equal to 1 by 9 divided by 1 by 3 and this is equal to 1 by 9 into 3 and this is equal to 1 by 3. We know that general term of a geometric progression is given by Tn is equal to A into R to the power n minus 1. Now in this question we have assumed that nth term that is Tn is equal to 1 by 1 9 6 8 3. Now Tn is equal to A into R to the power n minus 1. So this implies A into R to the power n minus 1 is equal to 1 by 1 9 6 8 3. A is equal to 1 by 3 and R is also equal to 1 by 3 by substituting the values of A and R we get 1 by 3 into 1 by 3 to the power n minus 1 is equal to 1 by 1 9 6 8 3. Now this implies 1 by 3 to the power n is equal to 1 by 1 9 6 8 3 and this implies 1 by 3 to the power n is equal to we can write 1 by 1 9 6 8 3 as 1 by 3 to the power 9. On comparing powers we find that n is equal to 9. Hence 1 by 1 9 6 8 3 is the 9th term the given GP. This is our required answer. So this completes the session. Bye and take care.