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In a geometric sequence, the term [tex]a_{n+1}[/tex] can be smaller than the term [tex]a_n[/tex].

A. True
B. False

Sagot :

In a geometric sequence, the term [tex]\( a_{n+1} \)[/tex] can indeed be smaller than the term [tex]\( a_n \)[/tex].

Here's a detailed explanation:

A geometric sequence is defined by the relationship between consecutive terms, where each term is obtained by multiplying the previous term by a constant called the common ratio, denoted as [tex]\( r \)[/tex].

Mathematically, this is expressed as:
[tex]\[ a_{n+1} = a_n \cdot r \][/tex]

For [tex]\( a_{n+1} \)[/tex] to be smaller than [tex]\( a_n \)[/tex], the common ratio [tex]\( r \)[/tex] must be less than 1. Specifically:
- If [tex]\( 0 < r < 1 \)[/tex], each subsequent term will be a fraction of the preceding term, making [tex]\( a_{n+1} \)[/tex] smaller than [tex]\( a_n \)[/tex].
- For instance, consider a sequence where the first term [tex]\( a_1 = 16 \)[/tex] and the common ratio [tex]\( r = 0.5 \)[/tex]. The sequence will be:
[tex]\[ 16, 8, 4, 2, 1, 0.5, \ldots \][/tex]
Here, each term is half the previous term, clearly demonstrating that [tex]\( a_{n+1} < a_n \)[/tex].

Therefore, based on this understanding, the statement is:

A. True