Third term = t3 = ar^2 = 444 eq. (1)
Seventh term = t7 = ar^6 = 7104 eq. (2)
By solving (1) and (2) we get,
ar^2 = 444
=> a = 444 / r^2 eq. (3)
And ar^6 = 7104
(444/r^2)r^6 = 7104
444 r^4 = 7104
r^4 = 7104/444
= 16
r2 = 4
r = 2
Substitute r value in (3)
a = 444 / r^2
= 444 / 2^2
= 444 / 4
= 111
Therefore a = 111 and r = 2
Therefore t6 = ar^5
= 111(2)^5
= 111(32)
= 3552.
<span>Therefore the 6th term in the geometric series is 3552.</span>
Answer:
5
Step-by-step explanation:
10% = 0.1
0.1*50 = 5
Ok so the average is 50
average of x number of things is
(add value of all things)/x=average
4 test scores needed
so divide by 4
3 test scores ar known
average is 50
x= missing score
(43+48+42+x)/4=50
times 4 both sides
133+x=200
minus 133 both sides
x=67
a score of 67 is needed