Question

If the third term is 3 of a geometric is 36 and the sixth term is 9/2, what is the explicit formula for the sequence ?

Answer 1

The formula will be `144(1/2)^(n - 1)` where `n ≥ 1`.

Explanation:

If `36` is the 3rd term, than the 4th term is `36r`, the 5th term is `36(r)(r) = 36r^2` and the 6th term is `36r(r)(r) = 36r^3`.

Thus:

`36r^3 = 9/2`

`r^3 = 9/72`

`r^3 = 1/8`

`r = 1/2`

The first term will be `144`, because the third term is `36`, so by the common ratio the second term is `36/(1/2) = 72` and the first is `72/(1/2) = 144`.

So the formula for the sequence is `144(1/2)^(n - 1)`, because the count starts at `n = 1` for term `1`.

Hopefully this helps!

Answer 2

`a_n = 144*(1/2)^(n-1)`

Explanation:

The general (`n^(th)`) term of a geometric sequence is given by:

`a_n = a_1*r^(n-1)`

Where `a_1` is the first term and (`r`) the common ratio

In this example we are given two terms, as follows:

`a_3 = 36 and a_6 = 9/2`

Applying the formula for the general term above:

`36 = a_1r^2` [A]

`9/2 = a_1r^5` [B]

[A] `-> r^2 = 36/a_1` [C]

[C] in [B] `-> 9/2 = a_1 * (36/a_1)^(5/2)`

Square both sides:

`81/4 = a_1^2 * (36/a_1)^(5)`

`81/4 = 36^5/a_1^3`

Expressing `a_1^3` in prime factors

`a_1^3 = (2*2*(3*3*2*2)^5)/(3*3)^2`

`= (2*2)^6*(3*3)^3`

`:. a_1 = (2*2)^2*(3*3)`

`= 16xx9 = 144`

`a_1 = 144` in [C] `-> r^2 = 36/144 = 1/4`

Hence, `r = +-sqrt(1/4) = =+-1/2`

Since `a_3 and a_6>0 -> r>0`

`:. r= 1/2`

Thus our general term for the sequence is:

`a_n = 144*(1/2)^(n-1)`