Question

6.1.27

Answer 1

Given that the nth rule for the sequence is $A(n)=8+(n-1)(12)$

We need to determine the seconds, fourth and eleventh terms of the sequence.

Second term:

The second term of the sequence can be determined by substituting n = 2, in $A(n)=8+(n-1)(12)$

Thus, we have;

$A(2)=8+(2-1)(12)$

Simplifying, we get;

$A(2)=8+(1)(12)$

$A(2)=8+12$

$A(2)=20$

Thus, the second term of the sequence is 20.

Fourth term:

The fourth term of the sequence can be determined by substituting n = 4 in $A(n)=8+(n-1)(12)$

Thus, we have;

$A(4)=8+(4-1)(12)$

Simplifying, we get;

$A(4)=8+(3)(12)$

$A(4)=8+36$

$A(4)=44$

Thus, the fourth term of the sequence is 44.

Eleventh term:

The eleventh term of the sequence can be determined by substituting n = 11  in $A(n)=8+(n-1)(12)$

Thus, we have;

$A(11)=8+(11-1)(12)$

Simplifying, we get;

$A(11)=8+(10)(12)$

$A(11)=8+120$

$A(11)=128$

Thus, the eleventh term of the sequence is 128.