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

Sagot :

ANSWER:

59049/512

EXPLANATION:

Given:

Common ratio(r) = 3/2

First term(a) = 3

Number of terms(n) = 10

To find:

The 10th term of the geometric sequence

We can go ahead and determine the 10th term of the sequence using the below formula and substituting the given values into it and evaluate;

[tex]\begin{gathered} a_n=ar^{n-1} \\ \\ a_{10}=3(\frac{3}{2})^{10-1} \\ \\ a_{10}=3(\frac{3}{2})^9 \\ \\ a_{10}=3(\frac{19683}{512}) \\ \\ a_{10}=\frac{59049}{512} \end{gathered}[/tex]

Therefore, the 10th term of the sequence is 59049/512