1. ## Geometric sequence

The fifth term of a geometric series is 5! and the sixth term is 6!. What is the fourth term?

Can someone post their answer and reasoning. I have my answer and think it's right but i really need e.c right now and want to make sure. Thanks!

2. Originally Posted by jarny
The fifth term of a geometric series is 5! and the sixth term is 6!. What is the fourth term?

Can someone post their answer and reasoning. I have my answer and think it's right but i really need e.c right now and want to make sure. Thanks!
For a geometric sequence {a_n}:
a_n = a_1*r^{n - 1}
where a_1 is the first term in the series and a_n is the nth term.

So
a_5 = 5! = 120 = a_1*r^4
a_6 = 6! = 720 = a_1*r^5

Two equations, two unknowns, we can solve this.

Divide a_6 by a_5:

a_6/a_5 = 6!/5! = 6 = (a_1*r^5)/(a_1*r^4) = r

Thus r = 6.

So
120 = a_1*6^4

Thus
a_1 = 120/6^4 = 20/6^3 = 20/216 = 5/54

So
a_n = (5/54)*6^{n - 1}

So
a_4 = (5/54)*6^{4 - 1} = (5/54)*6^3 = (5/54)*216 = 20

-Dan

3. got the same answer, thanks a lot

4. Hello, jarny!

The fifth term of a geometric series is 5! and the sixth term is 6!.
What is the fourth term?
Dan's solution is absolutely correct.
. . That's the approach I would have used.

But upon reflection, I realized that I let the factorials dazzle me.
. . There is a simpler solution.

We are given: .a
5 = 120 .and .a6 = 720

Then: .r .= .a
6/a5 .= .720/120 .= .6

. . The "rule" is multiply-by-six.

Therefore, the preceding term is: .a
4 = 20.

See? .We could have eyeballed the problem . . .

5. Originally Posted by Soroban

. . There is a simpler solution.
I always tell my students that I have a tendancy to make things harder than they have to be.

-Dan