Certainly, let's break down and solve the problem step by step.
Given:
- The first term of the sequence is [tex]\( a_1 = 2 \)[/tex]
- The sequence follows the rule [tex]\( a_{n+1} = 3a_n \)[/tex]
(i) Find the value of [tex]\( a_2 \)[/tex] :
To find [tex]\( a_2 \)[/tex], we use the rule [tex]\( a_{n+1} = 3a_n \)[/tex]:
[tex]\[ a_2 = 3a_1 \][/tex]
Since [tex]\( a_1 = 2 \)[/tex]:
[tex]\[ a_2 = 3 \times 2 = 6 \][/tex]
So, the value of [tex]\( a_2 \)[/tex] is [tex]\( 6 \)[/tex].
(ii) Compare [tex]\( a_2 \)[/tex] and [tex]\( a_3 \)[/tex] :
Next, we need to find [tex]\( a_3 \)[/tex]. Again, using the rule [tex]\( a_{n+1} = 3a_n \)[/tex]:
[tex]\[ a_3 = 3a_2 \][/tex]
We already found [tex]\( a_2 = 6 \)[/tex]:
[tex]\[ a_3 = 3 \times 6 = 18 \][/tex]
So, the value of [tex]\( a_3 \)[/tex] is [tex]\( 18 \)[/tex].
Now, to compare [tex]\( a_2 \)[/tex] and [tex]\( a_3 \)[/tex], we can simply look at their values:
- [tex]\( a_2 = 6 \)[/tex]
- [tex]\( a_3 = 18 \)[/tex]
Clearly, [tex]\( a_3 \)[/tex] is greater than [tex]\( a_2 \)[/tex]. To quantify the comparison:
[tex]\[ a_3 - a_2 = 18 - 6 = 12 \][/tex]
Therefore, [tex]\( a_3 \)[/tex] is greater than [tex]\( a_2 \)[/tex] by [tex]\( 12 \)[/tex].
Summary:
- [tex]\( a_2 = 6 \)[/tex]
- [tex]\( a_3 = 18 \)[/tex]
- [tex]\( a_3 \)[/tex] is greater than [tex]\( a_2 \)[/tex] by [tex]\( 12 \)[/tex].
These detailed steps confirm the values and comparison as requested.
