Easy
recall that
an=a1(r)^(n-1)
so
given 2nd and 5th term
we get
a2 and a5
so
a2=a1(r)^(2-1)=a1(r)^1=a1r
a5=a1(r)^(5-1)=a1(r)^4
also remember that
so
so r^3=-64
cube root
r=-4
so
a2=a1r=-8
a2=a1(-4)=-8
divide both sides by -4
a1=2
so
equation is
C isi the answer
recall that
an=a1(r)^(n-1)
so
given 2nd and 5th term
we get
a2 and a5
so
a2=a1(r)^(2-1)=a1(r)^1=a1r
a5=a1(r)^(5-1)=a1(r)^4
also remember that
so
so r^3=-64
cube root
r=-4
so
a2=a1r=-8
a2=a1(-4)=-8
divide both sides by -4
a1=2
so
equation is
C isi the answer