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## anonymous one year ago In a geometric sequence, u3 = 3, u7 = 48. Find r,u1, and u10

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1. anonymous

$a _{n}=ar ^{n-1}$

2. anonymous

So do I just work backwards or is there a different equation

3. freckles

$u_1=a \\ u_2=ar \\ u_3=ar^2 \\ ... \\ u_n=ar^{n-1} \\ \text{ you have } u_3=a r^{3-1}=ar^2=3 \\ \text{ and } u_7=ar^{7-1}=ar^6 =48 \text{ so this means } \frac{u_7}{u_3}=\frac{48}{3} \\ \text{ gives } \frac{ar^6}{ar^2}=\frac{48}{3} \\ \text{ solve for } r$

4. anonymous

thank you

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