A geometric sequence is defined recursively by an = 4an-1. The 1st term of the sequence is 0.5. Which of the following is the explicit formula for the nth term of the sequence?
\(\left\{\begin{array}{ccc}a_1=0.5\\a_n=4a_{n-1}\end{array}\right\\\\a_1=0.5\\a_2=0.5\cdot4=2\\a_3=2\cdot4=8\\a_4=8\cdot4=32\\a_5=32\cdot4=128\\\\a_n=a_1r^{n-1}\\\\r=a_2:a_1\to r=2:0.5=4\\\\\boxed{a_n=0.5\cdot4^{n-1}}\)
Answer:
\(a_n = 0.5 \cdot 4^{n-1}\)
Step-by-step explanation:
The explicit formula for the nth term of the geometric sequence is given by:
\(a_n = a_1 \cdot r^{n-1}\) ....[1]
As per the statement:
A geometric sequence is defined recursively by:
\(a_n = 4 \cdot a_{n-1}\) ....[2]
The 1st term of the sequence is 0.5
⇒\(a_1 = 0.5\)
We know that:
the recursive formula for geometric sequence is given by:
\(a_n = r \cdot a_{n-1}\) where, r is the common ratio
Compare with [2] we have;
r = 4
Substitute the values of r = 4 and \(a_1 = 0.5\) in [1] we have;
\(a_n = 0.5 \cdot 4^{n-1}\)
Therefore, the explicit formula for the nth term of the sequence is, \(a_n = 0.5 \cdot 4^{n-1}\)