Question

What is the 7th term of the geometric sequence where a1 = 625 and a3 = 25?

Answer 1

you've not provided enough information. What's the formula for the geometric sequence?

Answer 2

idk the formula :(

Answer 3

a3 = sqrt of a1
a5 = sqrt of a3
a7 = sqrt of a5

Answer 4

my options are
0.0016
0.008
0.04
0.2

Answer 5

@SomeoneYouUsedToKnow is that a geometric sequence? It's been a while for me, but I think a geometric sequence must have each successive term as a constant multiple of the previous term... So I don't think that taking square roots is an option... I might be wrong...

Answer 6

It's a good guess of me because 25² = 625 thats the only hidden clue i see.
In my opinion, i am correct. And yes its a geometric sequence.

Answer 7

Oh yeah, my mistake.

Answer 8

Happens :)

Answer 9

In a geometric series, the next element is obtained by multplying the previous by a constant. So a2=c*625. Then a3=c^2*625=25. This tells you that c=1/5. Then a3=25. So a7=c^4*25=(1/5)^4*25=0.04

Answer 10

For that last line, a3=25 so a4=c*25 and a5=c^2*25 and a6=c^3*25 and a7=c^4*25.

Answer 11

the thing is that you said in the statement that a1 = 625 and a3 = 25. so it's wrong what you try to explain :/

Answer 12

\[
r= \frac 1 5\\
a_7 = a_1 r^6= \frac 1 {25}
\]

Answer 13

yeah the distance between two articulars is 0,2 thats correct :)

Answer 14

@SomeoneYouUsedToKnow are you referring to what I said? I don't understand...

Answer 15

you just misspelled one number. you are right

Answer 16

I don't think I did.