Question

Find the 6th term of a geometric sequence t3 = 444 and t7 = 7104. So I have to find r. But is this right: 7104 = 444r^4 r^4 = 16 r = 2 Or should it be r^3? I'm never sure if the power is = the number of terms missing or something completely unrelated.

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.