 Hi and welcome to the session. I'm Kanika and I'm going to help you to solve the following question. The question says if the fourth, tenth and sixteenth terms of a gpr, x, y and z respectively prove that x, y, z are in GP. Let's now begin with the solution. Let a be the first term r be the common ratio of a GP. Question, it is given that fourth term of GP is equal to x. That means t4 is equal to x and t4 is equal to x implies a into r to the power 3 is equal to x. It is also given in the question that tenth term of a GP is y. That means tta is equal to y and this implies a into r to the power 9 is equal to y. It is also given in the question that sixteenth term of a GP is equal to z. That means t16 is equal to z and t16 is equal to z implies a into r to the power 15 is equal to z. Now we have to prove that x by z are in GP. Since x is equal to a into r to the power 3, y is equal to a into r to the power 9 and z is equal to a into r to the power 15. Therefore x, y, z will be in GP if a into r to the power 3, a into r to the power 9 and a into r to the power 15 are in GP. So x, y, z will be in GP if a into r to the power 3, a into r to the power 9 and a into r to the power 15 are in GP. Now this will be in GP if ratio of second and fourth term that is a into r to the power 9 upon a into r to the power 3 is equal to ratio of third and second term that is a into r to the power 15 upon a into r to the power 9. Now this implies r to the power 6 is equal to r to the power 6 and this is true. Therefore a into r to the power 3, a into r to the power 9 and a into r to the power 15 is a geometric progression and hence y and z GP. This completes this session. Bye and take care.