Question

The first term of geometric term is 6 and comman ratio is 3 what is the 6th term??

Answer 1

The `6^"th"` term is `a_6 = 1458`.

Explanation:

The formula for the `n^"th"` term in a geometric sequence is

`a_n = a_1xxr^(n-1)`

where

• `a_n` is the `n^"th"` term in the series,
• `a_1` is the `1^"st"` term in the series, and
• `r` is the common ratio between terms.

We know `a_1 = 6` and `r = 3`, and we want to find `a_6`. So, we can plug these into the formula above:

`a_n = a_1xxr^(n-1)`
`a_6 = 6 xx 3^(6-1)`
`color(white)(a_6)=6 xx 3^5`
`color(white)(a_6)=6 xx 243`
`color(white)(a_6)=1458`

Why does this formula work?

Having a common ratio `r` between terms means that each term is `r` times as big as the one before it.

For example, the `2^"nd"` term is `r` times as big as the `1^"st"`:

`a_2 = a_1 xx r`

And the `3^"rd"` term is `r` times as big as the `2^"nd"`:

`a_3 = a_2 xx r`

But wait—`a_2` is equal to `a_1 xx r`, so we can plug that in like this:

`a_3 = (a_1 xx r) xx r`
`color(white)(a_3)= a_1 xx r^2`

Since the next term will again be `r` times as big, all we are really doing is multiplying by `r` for every new term in the sequence.

Thinking of it this way, our terms in the sequence become:

`ul(" "n" "|"  "1"      "2"       "3"        "4" "..."        "n"     ")`
`" "a_n"   "|"  "a" "ar" "ar^2" "ar^3"   " ... " "ar^(n-1)`

where `a = a_1` is the first term in the sequence.

This is where the formula `a_n = ar^(n-1)` comes from.