Arthur D. answered 04/08/20
Forty Year Educator: Classroom, Summer School, Substitute, Tutor
f(t)=a1+d(n-1) where a1=first term, d=common difference, n=the number of the term
f(t)=4+(5)(n-1)
f(99)=4+(5)(99-1)
f(99)=4+(5)(98)
f(99)=4+490
f(99)=494
t(4)=12 and t(10)=48
we have to add some number 6 times because we are missing...
t(5), t(6), t(7), t(8), t(9), and we have to add this number to t(9) to get t(10)
12+6d=48
6d=36
d=6, the common difference
now go backwards to get t(1), the first term
12-6=6, 6-6=0, 0-6=-6
-6 is the first term
-6, 0, 6, 12, 18, 24, 30, 36, 42, 48, ... is the sequence
t(x)=-6+6(n-1) is the formula
t(18)=-6+6(18-1)
t(18)=-6+6(17)
t(18)=-6+102
t(18)=96