Answer:
see explanation
Step-by-step explanation:
The nth term of a geometric progression is
[tex]a_{n}[/tex] = a₁[tex]r^{n-1}[/tex]
where a₁ is the first term and r the common ratio
Given a₅ = 162 and a₈ = 4374 , then
a₁[tex]r^{4}[/tex] = 162 → (1)
a₁[tex]r^{7}[/tex] = 4374 → (2)
Divide (2) by (1)
[tex]\frac{a_{1}r^{7} }{a_{1}r^{4} }[/tex] = [tex]\frac{4374}{162}[/tex]
r³ = 27
r = [tex]\sqrt[3]{27}[/tex] = 3
Substitute r = 3 into (1) and solve for a₁
a₁[tex](3)^{4}[/tex] = 162
81a₁ = 162
a₁ = [tex]\frac{162}{81}[/tex] = 2
Then
a₂ = a₁ × 3 = 2 × 3 = 6
a₃ = a₂ × 3 = 6 × 3 = 18
The first 3 terms are 2, 6, 18
(ii)
The sum to n terms of a geometric progression is
[tex]S_{n}[/tex] = [tex]\frac{a_{1}(r^{n}-1) }{r-1}[/tex] , then
[tex]S_{10}[/tex] = [tex]\frac{2(3^{10}-1) }{3-1}[/tex]
= [tex]\frac{2(59049-1)}{2}[/tex]
= 59049 - 1
= 59048
