Answer:
Option A is correct.
[tex]a_1 = 12[/tex]
[tex]a_n = a_{n-1}+4[/tex]
Step-by-step explanation:
For a sequence [tex]a_1, a_2, a_3, . . . , a_n, . . .[/tex]
A recursive formula states that it is a formula that requires the computation of all previous terms in order to find the value of [tex]a_n[/tex] i,e it is given by;
[tex]a_n = a_{n-1}+d[/tex] ......[1]
Given the sequence : 12, 16, 20, 24, 28, .......
here,
First term = [tex]a_1 = 12[/tex]
[tex]a_2 = 16[/tex]
[tex]a_3 = 20[/tex]
[tex]a_4 = 24[/tex] and so on....
Now, find the common difference(d);
Common difference states that it is the difference between two numbers in an arithmetic sequence
Therefore, from the given sequence ;
d = 4
Since,
16 -12 = 4,
20-16 =4,
24 -20 = 4 and so on.....
Now, substitute the value of [tex]a_1 = 12[/tex] and d =4 in [1] ; we get
[tex]a_n = a_{n-1}+4[/tex]
Therefore, we have;
[tex]\left\{{{a_1=12} \atop {a_n =a_{n-1}+4}} \right[/tex]
