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!
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
Hello, jarny!
Dan's solution is absolutely correct.The fifth term of a geometric series is 5! and the sixth term is 6!.
What is the fourth term?
. . 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: .a5 = 120 .and .a6 = 720
Then: .r .= .a6/a5 .= .720/120 .= .6
. . The "rule" is multiply-by-six.
Therefore, the preceding term is: .a4 = 20.
See? .We could have eyeballed the problem . . .