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

"image"

step-by-step explanation:

in most of these problems, the notation is that the point with a "prime" is the image of the point without. that is, b' is the image of b.

here, there is no dilation or reflection. there is simply translation (with no direction shown). which figure you call the "preimage" and which one you call the "image" is up to you. as a matter of convention, the naming is as described above.