Question

Identify the 42nd term of an arithmetic sequence where a1 = -12 and a27 = 66

Answer 1

The arithmetic formula is expressed as an = a1 + d *(n-1)where n is an integer. Substituting from the given a1 = -12 and a27 = 66, 66 = -12 + d *(27-1). hence  , d is equal to 3. a42 thus using the formula is equal to 111. The final answer to this problem is 111.

Answer 2

Answer:

42nd term is 111

Step-by-step explanation:

$a_1=-12,a_{27}=66$

Now, Using formula of nth term of an A.P. we have :

$a_n=a+(n-1)\times d\\\text{Take n=27}\\\implies a_{27}=-12+(27-1)\times d\\\implies 66=-12+26\cdot d\\\implies 26\cdot d = 78\\\implies d = 3$

Now, first term a = -12 , n = 42 and d = 3

So, 42nd term is :

$a_{42}=-12+(42-1)\times 3\\\implies a_{42}=-12+41\times 3\\\implies a_{42}=-12+123\\\bf\implies a_{42}=111$

Hence, 42nd term is 111