Identify the 42nd term of an arithmetic sequence where a1 = −12 and a27 = 66. 70 72 111 114

Answer:111Step-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.