 Hi and welcome to the session. Let us discuss the following question. The question says which term of the following sequence is first part is 2 to root 2 4 and so on is 128. Let's now begin with the solution. Given geometric progression is and so on. Let let's term of the given GP 128. Now in the given GP first term that is a is equal to 2 and the common ratio that is R is equal to root 2. We know that we can obtain R by dividing any term by its preceding term. A general term of a geometric progression is given by Tn is equal to a into R to the power n minus 1. Now here we have assumed that and that's term of the given GPS 128. So this implies Tn is equal to 128. Now Tn is equal to a into R to the power n minus 1. So a into R to the power n minus 1 is equal to 128. We know that a is equal to 2 and R is equal to root 2. So by substituting the values of a and R we get 2 into root 2 to the power n minus 1. So 2 into root 2 to the power n minus 1 is equal to 128. Now this implies into we can write root 2 as 2 to the power n minus 1 by 2. So 2 into 2 to the power n minus 1 by 2 is equal to 128 and this implies to the power n minus 1 by 2 is equal to 64 and this implies 2 to the power n minus 1 by 2 is equal to we can write 64 as 2 to the power 6. Now on comparing exponents we find that n minus 1 by 2 is equal to 6 and this implies n minus 1 is equal to 12 and this implies n is equal to 13. Hence 128 is the thirteenth term of the GP. This is our required answer. So this completes the session. Bye and take care.