Answer:
a. T1 = -4 , T2 = -4/3 , T3 = -4/9 , T4 = -4/27
b. It is converge
c. The sum to ∞ = -6
Step-by-step explanation:
a.
∵ Tn = -4(1/3)^(n-1)
∵ The lower n = 1
∵ The geometric series Tn = a(r)^(n-1)
∴ T1 = a , T2 = ar , T3 = ar² , T4 = ar³
∴ a = -4 , r = 1/3
∴ T1 = -4
∴ T2 = -4(1/3) = -4/3
∴ T3 = -4(1/3)² = -4/9
∴ T4 = -4(1/3)³ = -4/27
b.
r = 1/3
∴ -1 < r < 1
∴ It is converge because the value of 1/3 when n is a very large
number will approach to zero
c.
∵ The sum of the geometric series = a(1-(r)^n)/1-r
∵ r^n ≅ 0 when n is a very large number
∴ The sum to ∞ = a/1 - r = -4/(1 - 1/3) = -4/(2/3) = -6
