17 | September | 2018 | Gavin's Blog

To find a term in a geometric sequence you must use the formula

$t_{n}$ = a $(r)^{(n-1)}$

to give an example here is my pattern -3, 9, -27, 81…
to find $t_{10}$ follow these steps….

First, $(r)^{(n-1)}$ becomes $(r)^{(10-1)}$ wich is $(r)^{9}$ leaving you with…

$t_{10}$ = a $(r)^{9}$

Then you want to add in (a), replace (a) with -3 because it’s your first term. making your equation look like…

$t_{10}$ = -3 $(r)^{9}$

once you have used all the given numbers you have. find the common ratio (r). if you follow the pattern of the sequence you will notice it is being multiplied by (-3) each time. so replace (r) with (-3)

$t_{10}$ = -3 $(-3)^{9}$

$(-3)^{9}$ is -19683

-3 x -19683 is 59,049

therefore $t_{10}$ = 59,049

– $Part 2$
to find the Geometric series you need to use this formula…

$S_{n}$ = $\frac{a (r^{n}-1)}{r-1}$

using the pattern above we will find the series of $S_{10}$ . first lets add in the values we were given straight of the bat. replace (a) with -3, and replace (n) with 10…