Question

Find the 13th term of the following sequence
-3,-18,-108...
a: 6*(-3)^13 c: -3*6^13
b: 6*(-3)^12 d: -3*6^12

Answer 1

it is a G.P.
a=-3
\[r=\frac{ -18 }{ -3 }=6\]
\[t _{n}=ar ^{n-1}\]
put n=13 and get the solution.

Answer 2

So its -3*6^12?

Answer 3

tn = ar^(n-1)
t13 = -3* 6^(13 - 1) =
t13 = -3 * 6^12
yep..you got it