The nth term of the geometric sequence is: an=ar^(n-1) where a=first term r=common ratio n=nth term from the question: 120=ar(3-1) 120=ar^2 a=120/(r^2)....i also: 76.8=ar^(5-1) 76.8=ar^4 a=76.8/r^4.....i thus from i and ii 120/r^2=76.8/r^4 from above we can have: 120=76.8/r² 120r²=76.8 r²=76.8/120 r²=0.64 r=√0.64 r=0.8 hence: a=120/(0.64)=187.5 therefore the formula for the series will be: an=187.5r^0.8
<span>The nth term of the geometric sequence is:
an=ar^(n-1)
where a=first term
r=common ratio
n=nth term
from the question:
120=ar(3-1)
120=ar^2
a=120/(r^2)....i
also:
76.8=ar^(5-1)
76.8=ar^4
a=76.8/r^4.....i thus from i and ii
120/r^2=76.8/r^4
from above we can have:
120=76.8/r²
120r²=76.8
r²=76.8/120
r²=0.64
r=âš0.64
r=0.8
hence:
a=120/(0.64)=187.5
therefore the formula for the series will be:
an=187.5r^0.8</span>