maths help

This is a geometric progression

The common ratio is   -6/4  = -3/2

So....the nth term in the series can be fornd as

a_n =  a₁ (r)^{(n - 1)}    where  a₁  =   4  and r  = -3/2

So......the 4th term is

a₄  =  4(-3/2)^{(4 - 1)}  =   4(-3/2)³    =  -27/2  =  -13.5

The first two terms of a geometric progression are 4 and  –6.  What is the 4th term?   

a=4 and d=-6-4=-10

\(T_n=a+(n-1)d\\ T_4=4+(4-1)*(-10)\\ T_4=4+(3)*(-10)\\ T_4=4+-30\\ T_4=-26\\ \)