Answer:
111
Step-by-step explanation:
It is given that [tex]a_{1}=-12[/tex] and [tex]a_{27}=66[/tex], thus
[tex]a_{1}=-12[/tex] ,
[tex]a_{2}=a_{1}+d=-12+d[/tex],
[tex]a_{3}=a_{2}+d=-12+2d[/tex],
.........
[tex]a_{27}=-12+26d[/tex]
⇒[tex]66=-12+26d[/tex]
⇒[tex]78=26d[/tex]
⇒[tex]d=3[/tex]
therefore, the value of [tex]a_{42}[/tex] will be:
[tex]a_{42}=a_{1}+41d[/tex]
=[tex]-12+41(3)[/tex]
=[tex]111[/tex]
Thus, [tex]a_{42}=111[/tex]
Therefore, C is the correct option.
The 42nd term of the arithmetic sequence whose first term is -12 and the common difference is of 3 is 111.
An arithmetic sequence is the series of numbers where the difference between any two consecutive numbers of the series is the same. The nth term of any arithmetic sequence is given as,
[tex]a_n = a_1 + (n-1)d[/tex]where [tex]a_n[/tex] is the nth term, [tex]a_1[/tex] is the first term of the sequence, and d is a common difference.
We know the first term of the sequence that is [tex]a_1 = -12[/tex], we also know the 27th term of the sequence, therefore, the difference of the arithmetic sequence can be written as,
[tex]a_n = a_1 + (n-1)d\\\\66 = -12 + (27-1)d\\\\66+12=(26)d\\\\d = 3[/tex]
Now, the 42nd term of the arithmetic sequence can be written as
[tex]a_{42} = a_1 + (n-1)d\\\\a_{42}= -12 + (42-1)3\\\\a_{42} = 111[/tex]
Hence, the 42nd term of the arithmetic sequence is 111.
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