A(n)=−5+6(n−1)a, left parenthesis, n, right parenthesis, equals, minus, 5, plus, 6, left parenthesis, n, minus, 1, right parenthesis find the 12^\text{th}12 th 12, start superscript, t, h, end superscript term in the sequence. answer
Given that A(n)=-5+6(n-1)a. to get the 12th term we need to substitute the values as follows; n=12 thus A(12)=-5+6(12-1)a A(12)=-5+6(11)a A(12)=-5+66a A(12)=66a-5 The answer is A(12) is 66a-5
