Find the 8th term of an exponential sequence whose first term is 3 and whose common ratio is 2
Please explain this to me
To find the 8th term of an exponential sequence, we need to use the formula:
\[a_n = a_1 \times r^{(n-1)}\]
where:
- \(a_n\) is the nth term of the sequence
- \(a_1\) is the first term of the sequence
- \(r\) is the common ratio of the sequence
- \(n\) is the term number we want to find
Given that the first term (\(a_1\)) is 3, and the common ratio (\(r\)) is 2, we can substitute these values into the formula to find the 8th term:
\[a_8 = 3 \times 2^{(8-1)}\]
Now let's solve the equation:
\[a_8 = 3 \times 2^7\]
\[a_8 = 3 \times 128\]
\[a_8 = 384\]
Therefore, the 8th term of the exponential sequence is 384.
To find the 8th term of an exponential sequence, we need to use the formula for the nth term of an exponential sequence. The formula is:
a_n = a * r^(n-1)
Where:
a_n is the nth term of the sequence,
a is the first term of the sequence, and
r is the common ratio.
In this case, the first term (a) is 3 and the common ratio (r) is 2. We want to find the 8th term (a_8). Plugging the values into the formula, we have:
a_8 = 3 * 2^(8-1)
= 3 * 2^7
= 3 * 128
= 384
So, the 8th term of the exponential sequence is 384.
Recall that the nth term
an = a*r^(n-1)
so plug in your numbers...