Question

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

Answer 1

The formula for the nth term of a geometric sequence:
$a_n=a_1 \times r^{n-1}$
a₁ - the first term, r - the common ratio

$54, a_2, a_3, 128 \\ \\ a_1=54 \\ a_4=128 \\ \\ a_n=a_1 \times r^{n-1} \\ a_4=a_1 \times r^3 \\ 128=54 \times r^3 \\ \frac{128}{54}=r^3 \\ \frac{128 \div 2}{54 \div 2}=r^3 \\ \frac{64}{27}=r^3 \\ \sqrt[3]{\frac{64}{27}}=\sqrt[3]{r^3} \\ \frac{\sqrt[3]{64}}{\sqrt[3]{27}}=r \\ r=\frac{4}{3}$

$a_2=a_1 \times r= 54 \times \frac{4}{3}=18 \times 4=72 \\ a_3=a_2 \times r=72 \times \frac{4}{3}=24 \times 4=96 \\ \\ \boxed{a_2=72, a_3=96}$

Answer 2

$a=54 \\ \\ a2=54q \\ \\ a3=54q^{2} \\ \\ 128=54q^{3} \\ \\ q^{3}= \frac{128}{54}= \frac{64}{27} \\ \\ q= \frac{4}{3} \\ \\ a2= 54*\frac{4}{3}= 72 \\ \\ a3=54*( \frac{4}{3})^{2}=54* \frac{16}{9} =96$

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