The first term is 612.
The common ratio is 1.08 and
The recursive rule is [tex]a_{n} = a^{n-1} \times r[/tex]
Step-by-step explanation:
the question to the problem is to write the values of the first term, common ratio, and expression for the recursive rule.
The first term :
In geometric sequence, the first term is given as [tex]a_{1}[/tex].
⇒ [tex]a_{1} = 612[/tex]
Now, the geometric sequence follows as 612, 661, ........
The common ratio (r) :
It is the ratio between two consecutive numbers in the sequence.
Therefore, to determine the common ratio, you just divide the number from the number preceding it in the sequence.
⇒ r = 661 divided by 612
⇒ r = 1.08
To find the recursive rule :
A geometric series is of the form a,ar,ar2,ar3,ar4,ar5........
Here, first term [tex]a_{1} = a[/tex] and other terms are obtained by multiplying by r.
The recursive rule is of the form [tex]a_{n} = a^{n-1} \times r[/tex]
This is called recursive formula for geometric sequence.
We know that r = 1.08 and [tex]a_{1}[/tex] = 612.
To find the second term [tex]a_{2}[/tex], use the recursive rule [tex]a_{n} = a^{n-1} \times r[/tex]
⇒ [tex]a_{2} = a^{2-1}\times r[/tex]
⇒ [tex]a_{2} = a^{1}\times r[/tex]
⇒ [tex]a_{2} = 612\times 1.08[/tex]
⇒ [tex]a_{2} = 661[/tex]
