 Hi and welcome to the session. I am Asha and I am going to help you with the following question which says, a GP consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio. Let us now begin with the solution and let the GP be a, a r, a r square, a r cube. Now, the sum of these terms is equal to a into r raised to the power 2n minus 1 upon r minus 1. So, let this be equation number 1 and now let us find the sum of terms occupying odd spaces. So, the terms occupying odd spaces are a r square, a r raised to the power 4 and so on in terms. So, the sum of these n terms is equal to a. Now, the common difference is r square and the terms are n minus 1 upon r square minus 1 or we can say that as n is equal to a r raised to the power 2n minus 1 upon r square minus 1. So, let this be equation number 2. Given that the sum of all the terms is 5 times the sum of terms occupying odd places. So, its 2n is equal to 5 times of sn or a r raised to the power 2n minus 1 upon r minus 1 is equal to a into r raised to the power 2n minus 1 upon r square minus 1 into 5. Or we can say that a cancels out with a r raised to the power 2n minus 1 with r raised to the power 2n minus 1 and we have 1 upon r minus 1 in the left hand side. And on the right hand side we have 1 upon r minus 1 into r plus 1 into 5 or we can say that 1 is equal to 5 upon r plus 1 or r plus 1 is equal to 5 or r is equal to 4. Therefore, the common ratio is 4. So, this completes the solution. Take care and have a good day.