Question

How do I find a2 and a3 for the following geometric sequence? 54, a2, a3, 128

Answer 1

`a=54 \\ \\ a2=54q \\ \\ a3=54q^(2) \\ \\ 128=54q^(3) \\ \\ q^(3)= (128)/(54)= (64)/(27) \\ \\ q= (4)/(3) \\ \\ a2= 54*(4)/(3)= 72 \\ \\ a3=54*( (4)/(3))^(2)=54* (16)/(9) =96`

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Answer 2

The formula for the nth term of a geometric sequence:

`a_n=a_1 * r^(n-1)`
a₁ - the first term, r - the common ratio


`54, a_2, a_3, 128 \\ \\ a_1=54 \\ a_4=128 \\ \\ a_n=a_1 * r^(n-1) \\ a_4=a_1 * r^3 \\ 128=54 * r^3 \\ (128)/(54)=r^3 \\ (128 / 2)/(54 / 2)=r^3 \\ (64)/(27)=r^3 \\ \sqrt[3]{(64)/(27)}=\sqrt[3]{r^3} \\ \frac{\sqrt[3]{64}}{\sqrt[3]{27}}=r \\ r=(4)/(3)`


`a_2=a_1 * r= 54 * (4)/(3)=18 * 4=72 \\ a_3=a_2 * r=72 * (4)/(3)=24 * 4=96 \\ \\ \boxed{a_2=72, a_3=96}`