Formula for $n^(th)$ term of a sequence - multiplication

In order to predict the $n^(th)$ term of a sequence you will need to create a formula.

For sequence patterns of geometric progressions or geometric sequences (or multiplications) this is worked out by using the formula.

$ar^(n-1)$

Where

$a=$ first term

$r=$ the multiple

$n=n^(th)$ number

We can help you remember this as follows

Mnemonic to help you remember how to find the nth term using the multiplication formula

Arrrr $(ar)$ not one $(n-1)$ of us

$ar^(n-1)$

Example 1

We know the following sequence is a geometric sequence but what is the formula of the $n^(th)$ term and what is the $8^(th)$ term.

$2$	$6$	$18$	$54$	$162$
We can check that it is a multiple sequence by dividing each term by the previous term.
	$6/2=3$	$18/6=3$	$54/18=3$	$162/54=3$

We have a multiple of $3$

So the formula for this is

$ar^(n-1)$

So here

$a=$ first term $=2$

$r=$ the multiple $=3$

So the formula is

$2times3^(n-1)$

Now we need to check the formula is correct.

Try different values of $n$ in the formula $2times3^(n-1)$

$n=1$ term $=2times3^(1-1)=2times3^0=2times1=2$

$n=2$ term $=2times3^(2-1)=2times3^1=2times3=6$

$n=3$ term $=2times3^(3-1)=2times3^2=2times9=18$

$n=4$ term $=2times3^(4-1)=2times3^3=2times27=54$

This is correct

The 8th term would be

If $n=8$ then $2times3^(n-1)=2times3^(8-1)=2times3^7=4374$

The 8th term $=4,374$

Example 2

We know the following sequence is a geometric sequence but what is the formula of the $n^(th)$ term and what is the $8^(th)$ term.

$-2$	$4$	$-8$	$16$	$-32$	$64$
We can check that it is a multiple sequence by dividing each term by the previous term.
	$4/-2=-2$	$(-8)/4=-2$	$16/-8=-2$	$(-32)/16=-2$	$64/-32=-2$

We have a multiple of $-2$

So the formula for this is

$ar^(n-1)$

So here

$a=$ first term $=-2$

$r=$ the multiple $=-2$

So the formula is

$-2times(-2)^(n-1)$

Now we need to check the formula is correct.

Try different values of $n$ in the formula $-2times(-2)^(n-1)$

$n=1$ term $=-2times(-2)^(1-1)=-2times(-2)^0=-2times1=-2$

$n=2$ term $=-2times(-2)^(2-1)=-2times(-2)^1=-2times-2=4$

$n=3$ term $=-2times(-2)^(3-1)=-2times(-2)^2=-2times4=-8$

$n=4$ term $=-2times(-2)^(4-1)=-2times(-2)^3=-2times(-8)=16$

This is correct

The 8th term would be

If $n=8$ then $-2times(-2)^(n-1)=-2times(-2)^(8-1)=-2times(-2)^7$

$=-2times(-2xx-2xx-2xx-2xx-2xx-2xx-2)$

$=-2times(-128)=256$

Answer $=256$

Example 3

We know the following sequence is a geometric sequence but what is the formula for the $n^(th)$ term and what is the $8^(th)$ term.

$40$	$20$	$10$	$5$	$2.5$
We can check that it is a multiple sequence by dividing each term by the previous term.
	$20/40=0.5$	$10/20=0.5$	$5/10=0.5$	$2.5/5=0.5$

We have a multiple of $0.5$

So the formula for this is

$ar^(n-1)$

So here

$a=$ first term $=40$

$r=$ the multiple $=0.5$

So the formula is

$40times(0.5)^(n-1)$

Now we need to check the formula is correct.

Try different values of $n$ in the formula $40times(0.5)^(n-1)$

$n=1$ term $=40times(0.5)^(1-1)=40times0.5^0=40times1=40$

$n=2$ term $=40times(0.5)^(2-1)=40times0.5^1=40times0.5=20$

$n=3$ term $=40times(0.5)^(3-1)=40times0.5^2=40times0.5xx0.5=10$

$n=4$ term $=40times(0.5)^(4-1)=40times0.5^3=40times0.5xx0.5xx0.5=5$

This is correct

The 8th term would be

If $n=8$ then $40times(0.5)^(n-1)=40times(0.5)^(8-1)=40times0.5^7$

$=40times(0.5xx0.5xx0.5xx0.5xx0.5xx0.5xx0.5)$

$=0.3125$

Answer the 8th term $=0.3125$

Example 4

We know the following sequence is a geometric sequence but what is the formula of the $n^(th)$ term and what is the $8^(th)$ term.

$3$

$3sqrt8$

$24$

$24sqrt8$

We change to

$3$	$3sqrt8$	$3sqrt8sqrt8$	$3sqrt8sqrt8sqrt8$
We can check that it is a multiple sequence by dividing each term by the previous term.
	$(3sqrt8)/3=sqrt8$	$(3sqrt8sqrt8)/(3sqrt8)=sqrt8$	$(3sqrt8sqrt8sqrt8)/(3sqrt8sqrt8)=sqrt8$

We have a multiple of $sqrt8$

So the formula for this is

$ar^(n-1)$

So here

$a=$ first term $=3$

$r=$ the multiple $=sqrt8$

So the formula is

$3sqrt8^(n-1)$

Now we need to check the formula is correct.

Try different values of $n$ in the formula $3sqrt8^(n-1)$

$n=1$ term $=3sqrt8^(1-1)=3sqrt8^0=3times1=3$

$n=2$ term $=3sqrt8^(2-1)=3sqrt8^1=3sqrt8$

$n=3$ term $=3sqrt8^(3-1)=3sqrt8^2=3sqrt8sqrt8$

$n=4$ term $=3sqrt8^(4-1)=3sqrt8^3=3sqrt8sqrt8sqrt8$

This is correct

The 8th term would be

Finding the 8th term the long way around

Answer the 8th term $=1536sqrt8$

Formula for nthn^(th) term of a sequence - multiplication

Formula for $n^(th)$ term of a sequence - multiplication