If the 5th term of a geometric progression is 162 and the 8th term is 4374, find the (i) 1st three terms of the sequence; (ii) sum of the first 10 terms​

see explanation

Step-by-step explanation:

The nth term of a geometric progression is

= a₁

where a₁  is the first term and r the common ratio

Given a₅ = 162 and a₈ = 4374 , then

a₁ = 162 → (1)

a₁ = 4374 → (2)

Divide (2) by (1)

=

r³ = 27

r = = 3

Substitute r = 3 into (1) and solve for a₁

a₁ = 162

81a₁ = 162

a₁ = = 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

= , then

=

     =

     = 59049 - 1

     = 59048

the answer to this is a. a^2 + b^2 = c^2

22, 28, 34, 40, 46

b)   f(n) = 6n + 16

n = what is the number of the row ( 22 would be row 1) so if we replaced n with 1, we would get 22

c)   the 14th row will have 100 seats

(1)   (2)   (3)   (4)   (5)   (6)   (7)   (8)   (9)   (10) (11) (12) (13) (14)

22 | 28 | 34 | 40 | 46 | 52 | 58 | 64 | 70 | 76 | 82 | 88 | 94 | 100