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find the 10th term of the geometric sequence whose common ratio is 2/3 and whose first term is 3.

Sagot :

Evgeniyleviidn

Answer:

[tex]\frac{512}{6561}[/tex]

Step-by-step explanation:

1) if according to the condition a₁=3 and r=2/3, then a₁₀ can be calculated as:

a₁₀=a₁*r⁹;

2) according  to the rule above:

a₁₀=3*(2/3)⁹=512*3/(6561*3)=512/6561.

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