Question

Question #730a0

Answer 1

`27`

Explanation:

Denote the `n`:th element in the sequence `a_n = bx^(n-1)`.
Then `a_3=bx^2` and `a_6 = bx^5`. Knowing that `a_3 = 3` and `a_6=1/9`, we get that
`3 = b*x^2`
`1/9 = b*x^5 = (bx^2)x^3`
Substituting the first line into the second we get that
`1/9 = 3x^3`, which gives that
`x = 1/3`.
Inserting into the first equation, we get that
`b = 3/(x^2) = 27`.
Thus the first element `a_1 = bx^(1-1) = b = 27`.

Answer 2

`27.`

Explanation:

Let the Geom. Seq. be `a,ar, ar^2,...,ar^(n-1),...`

By what is given, `ar^2=3, and, ar^5=1/9.`

`:. (ar^5)/(ar^2)=1/9-:3`.

`:. r^3=1/27`

`:. r=1/3.`

Now, `ar^2=3, r=1/3 rArr a=3-:r^2=3-:(1/3)^2=3-:1/9=27.`

`:." The First term is "27.`