Question

Find the 6th term of a geometric sequence t3=444 t7=7104

Answer 1

Third term = t3 = ar^2 = 444           eq. (1)

Seventh term = t7 = ar^6 = 7104         eq. (2)

By solving (1) and (2) we get,

              ar^2 = 444    

                => a = 444 / r^2       eq. (3)

And  ar^6 = 7104

 (444/r^2)r^6 = 7104

 444 r^4 = 7104

 r^4 = 7104/444

            = 16

 r2 = 4

 r = 2

Substitute r value in (3)

                         a = 444 / r^2

                             = 444  / 2^2

                             = 444 / 4

                              = 111

Therefore a = 111 and r = 2

Therefore t6 = ar^5

                       = 111(2)^5

                       = 111(32)

                       = 3552.

Therefore the 6th term in the geometric series is 3552.