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Find the 7th term of the geometric sequence whose common ratio is 2/3 and whose first term is 4
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$$a_{7}=\frac{256}{729}$$ or $$a_{7}=0.35$$

Step-by-step explanation:

$$a_{n} =a_{1}* r^{n-1}$$

$$a_{7}=4*\frac{2}{3}^{7-1}$$

$$a_{7}=4*\frac{2}{3}^{6}$$

$$a_{7}=4*\frac{64}{729}$$

$$a_{7}=\frac{256}{729}$$

or

$$a_{7}=0.35$$
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