Question

Find the (2n)th term of the GP: -1/3, 1, -3, 9, ...

Answer 1

@ganeshie8

Answer 2

should be pretty straightforward :
\[\large \rm a_{2n} = a_1r^{2n-1}\]

Answer 3

?

Answer 4

so is it 3^(2n-2)

Answer 5

how ?

Answer 6

it should be \[\large \rm a_{2n} = \frac{-1}{3}(-3)^{2n-1}\]

Answer 7

right ?

Answer 8

yes

Answer 9

it simplifies to
\[\large (-3)^{2n-2}\]