# Geometric progression

just a check of my work please.

I have to write an expression for the nth term of this geometric sequence.

a1=100
a2=106
a3=112.36

I've worked out the ratio to be r=1.06

I am using the formula un=ar(n-1)

so the expression i have come up with is un=100(1.06)(n-1)

Is this correct? I have checked and it does work. I tried multiplying 100(1.06) on its own first to give me the formula
106(n-1)
but that doesnt work....hence me doubting myself.

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tiny-tim
Homework Helper
hi ghostbuster25! … so the expression i have come up with is un=100(1.06)(n-1)

Is this correct? I have checked and it does work. I tried multiplying 100(1.06) on its own first to give me the formula
106(n-1)
but that doesnt work....hence me doubting myself.
yup, un=100(1.06)(n-1) is fine (i don't understand how you got 106(n-1) )

Mentallic
Homework Helper
ghostbuster, if what you were trying to do is convert $$100(1.06)^{n-1}$$ into $$106^{n-2}$$, this is wrong because what you should be doing is taking out the factor of 1.06 to give $$100(1.06)(1.06)^{(n-1)-1}=106(1.06)^{n-2}$$.

106n grows much faster than 1.06n.