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Answer:
(c) 3360
Step-by-step explanation:
Given
The above arithmetic series where:
[tex]T_1 = 12[/tex] --- first term
[tex]d = 6[/tex] --- the common difference (18 - 12 = 6)
[tex]T_n = 198[/tex] --- The last term
Required
The value of the series
First, we calculate n using:
[tex]T_n = T_1 + (n - 1)d[/tex]
This gives:
[tex]198 = 12 + (n - 1)*6[/tex]
Collect like terms
[tex]198 - 12 = (n - 1)*6[/tex]
[tex]186 = (n - 1)*6[/tex]
Divide both sides by 6
[tex]31 = (n - 1)[/tex]
Make n the subject
[tex]n = 31 + 1[/tex]
[tex]n=32[/tex]
The sum of the series is:
[tex]S_n= \frac{n}{2}(T_1 + T_n)[/tex]
So, we have:
[tex]S_n= \frac{32}{2}(12 + 198)[/tex]
[tex]S_n= \frac{32}{2}*210[/tex]
[tex]S_n= 16*210[/tex]
[tex]S_n= 3360[/tex]