Question

What is the 9th term of the geometric sequence with `a_2 = .3` and `a_6 = .00003`?

Answer 1

`T_9 = 3` x `0.1^8 = 3` x `10^-8`

Explanation:

Each term in a GP has the general form `T_n = ar^(n-1)`

If you have the values of any two terms in a GP, a very easy way to find the general term is to divide the two terms - their formulae and their values. Divide the term further along the sequence by any one before it.

`T_6/T_2` = `(ar^6)/(ar^2)` = `0.00003/0.3`

The factors `a` will cancel, using the division law of indices and simplifying the fraction gives:

`r^4 = 1 xx 10^-4 = 0.0001`

Now find the 4th root to find the value of `r`
`root(4) 0.0001 = 0.1`

The value of the first term, `a` can be found by dividing the second term by `r`. This gives `a = 0.3/0.1 = 3`

We now have a value for `a` and for `r`, so it is an easy matter to find the value for the 9th term.

`T_9 = ar^(n-1) = 3` x `0.1^8 = 3` x `10^-8`