 Hi and welcome to the session. Let us discuss the following question. The question says find the sum to indicated number of terms in each of the geometric progressions in exercises 7 to 10. Given geometric progression is root 7, root 21, 3 root 7, so on up to n tops. Before solving this question, we should know that sum of first n terms of Gp that is Sn is given by a into r to the power n minus 1 upon r minus 1 if r is greater than 1. a is the first term of the Gp, r is the common ratio, n is the number of terms. The knowledge of this formula is the key idea in this question. Now begin with the solution. Given Gp is root 7, root 21, 3 root 7, so on up to n terms. We have to find sum of n terms of this Gp. Now here the first term that is a is equal to root 7, common ratio that is r is equal to root 3 and root 3 is greater than 1. We have learned in the key idea that if r is greater than 1, then sum of n terms of Gp is given by a into r to the power n minus 1 upon r minus 1. Now substitute the value of a and r in this formula. By substituting the values, we get root 7 into root 3 to the power n minus 1 upon root 3 minus 1. On rationalizing it, we get root 7 into root 3 to the power n minus 1 into root 3 plus 1 upon root 3 minus 1 into root 3 plus 1 equal to root 7 into root 3 plus 1 into we can write root 3 to the power n as 3 to the power n by 2 minus 1 upon root 3 minus 1 into root 3 plus 1 is equal to root 3 square minus 1 square by using the identity of a square minus b square and this is equal to 3 minus 1 and this is equal to 2. So we have 2 in the denominator. Hence the required sum is root 7 by 2 into root 3 plus 1 into 3 to the power n by 2 minus 1. This is our required answer. So this completes the session. Bye and take care.