Question

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

Answer 1

Answer:

111

Step-by-step explanation:

It is given that $a_{1}=-12$ and $a_{27}=66$, thus

$a_{1}=-12$ ,

$a_{2}=a_{1}+d=-12+d$,

$a_{3}=a_{2}+d=-12+2d$,

.........

$a_{27}=-12+26d$

⇒$66=-12+26d$

⇒$78=26d$

⇒$d=3$

therefore, the value of $a_{42}$ will be:

$a_{42}=a_{1}+41d$

=$-12+41(3)$

=$111$

Thus, $a_{42}=111$

Therefore, C is the correct option.

Answer 2

The 42nd term is 111.