Question

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

Answer 1

42nd term is 111

Explanation:

`a_1=-12,a_(27)=66`

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

`a_n=a+(n-1)* d\\\text{Take n=27}\\\implies a_(27)=-12+(27-1)* 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)* 3\\\implies a_(42)=-12+41* 3\\\implies a_(42)=-12+123\\\bf\implies a_(42)=111`

Hence, 42nd term is 111

Answer 2

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.