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

you got the value of r correctly. but the method you used seems wrong

Answer 2

im just lil curious... can you pls explain how you put this :-

7104 = 444r^4

Answer 3

I have no idea what formula I used I just followed the example of another problem (from the pic)

Answer 4

and since they didn't explain how they got to \[r^{2}\] I just assumed it was the number of terms or something. But I'm not sure

Answer 5

t3 = 444 and t7 = 7104

you could figure it like this: multiply a term by r to get the next term:
t4= t3*r
t5= t4*4 = (t3*r)*r= t3*r^2
t6= t5*r= t3*r^3
t7 = t6*r= t3*r^4

to get from t3 to t4 you multiply by r once
to get from t3 to t7 you multiply by r four times, r^4

Answer 6

let a be the first term ,r the common ratio.
tn=ar^(n-1)
t3=ar^2=444
t7=ar^6=7104
divide
\[\frac{ ar ^{6} }{ar ^{2} }=\frac{ 7104 }{444 }=16\]
r^4=2^4
r=2
a*2^2=444
a=444/4 =111
t6=ar^5=111*2^5=111*32=3552
or t6*r=t7
t6=t7/r=7104/2=3552

Answer 7

phi, that makes it very clear. Thanks a lot!