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Sagot :
Given the sequence with the first term as 5, and each subsequent term being 3 times the preceding term, we can establish that this sequence is geometric with the first term ([tex]\(a\)[/tex]) equal to 5 and the common ratio ([tex]\(r\)[/tex]) equal to 3.
For a geometric sequence, the [tex]\(n\)[/tex]-th term is given by:
[tex]\[ a_n = a \cdot r^{n-1} \][/tex]
Here, [tex]\(a = 5\)[/tex] and [tex]\(r = 3\)[/tex]. Therefore, the [tex]\(m\)[/tex]-th term of the sequence, [tex]\(x\)[/tex], can be written as:
[tex]\[ x = 5 \cdot 3^{m-1} \][/tex]
Thus, the correct equation for [tex]\(x\)[/tex] in terms of [tex]\(m\)[/tex] is:
(B) [tex]\(\boxed{x = 5 \cdot 3^{m-1}}\)[/tex]
For a geometric sequence, the [tex]\(n\)[/tex]-th term is given by:
[tex]\[ a_n = a \cdot r^{n-1} \][/tex]
Here, [tex]\(a = 5\)[/tex] and [tex]\(r = 3\)[/tex]. Therefore, the [tex]\(m\)[/tex]-th term of the sequence, [tex]\(x\)[/tex], can be written as:
[tex]\[ x = 5 \cdot 3^{m-1} \][/tex]
Thus, the correct equation for [tex]\(x\)[/tex] in terms of [tex]\(m\)[/tex] is:
(B) [tex]\(\boxed{x = 5 \cdot 3^{m-1}}\)[/tex]
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