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>
x³ + 2x² - x - 2 < 0
x³ - x + 2x² - 2 < 0
x(x² - 1) - 2(x² - 1) < 0
(x - 2)(x² - 1) < 0
(x - 2)(x - 1)(x + 1) < 0
by chacking we get the solution
x ∈ (-∞,-1) ∪ (-1,1)
Circle F, diameter is 16 and the radius is half of the diameter
Answer:
9
Step-by-step explanation:
