 Hi and welcome to the session. I'm Kanika and I'm going to help you to solve the following question. The question says the 4 term of a GP is square of its second term and the first term is minus 3. Determine its seventh term. Let's now begin with this illusion. Let A be the first term be the common ratio of a GP. The question it is given that the 4 term of a GP that is T4 is equal to square of its second term that is T2 whole square. Now we know that in its term of a geometric progression is given by Tn is equal to A into r to the power n minus 1. So fourth term that is T4 is equal to A into r to the power 4 minus 1 and this is equal to A into r to the power 3 and the second term that is T2 is equal to A into r to the power 2 minus 1 and this is equal to A into r. T4 is equal to T2 whole square implies A r q is equal to A r whole square and this implies A r q is equal to A square r square and this implies r is equal to A now in the question we are given that A is equal to minus 3 right so this implies r is equal to minus 3. The value of r so we can now easily find the seventh term now seventh term is equal to A into r to the power 7 minus 1 this is equal to A into r to the power 6 substitute the value of A and r. A is equal to minus 3 and r is also equal to minus 3 and this is equal to minus 3 to the power 7 and this is equal to minus 2187 hence the required seventh term that is T7 is equal to minus 2187 this is our required answer so this completes this session bye and take care