 Hi, welcome to the session. Let us discuss the following question. The question says, how many terms of GP3, 3 squared, 3 cubed, so on, are needed to give sum 120? Let's now begin with this solution. In this question, we have to find number of terms of GP3, 3 squared, 3 cubed, and so on, which are needed to give sum 120. So let n be the number of terms needed. Given GP is 3, 3 squared, 3 cubed, and so on. Now here, the first term, that is, a is equal to 3, and the common ratio, that is, r is also equal to 3. We know that if r is greater than 1, then sum to n terms of GP, that is, Sn is equal to a into r to the power n minus 1 upon r minus 1. Now here, sum to n terms of the GP, that is, Sn is given as 120. So now, we will substitute the value of a, r, and Sn in this formula. By substituting the values, we get 120 is equal to 3 into 3 to the power n minus 1 upon 3 minus 1. Now this implies 120 is equal to 3 by 2 into 3 to the power n minus 1. And this implies 240 by 3 is equal to 3 to the power n minus 1. This implies 80 is equal to 3 to the power n minus 1. And this implies 81 is equal to 3 to the power n. And this implies we can write 81 as 3 to the power 4. So we have 3 to the power 4 is equal to 3 to the power n. Since the base is same, therefore, on comparing powers, we get n is equal to 4. Hence the required value of n is 4. This completes the session. Bye and take care.