Question

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​

Answer 1

Answer:

see explanation

Step-by-step explanation:

The nth term of a geometric progression is

$a_{n}$ = a₁$r^{n-1}$

where a₁  is the first term and r the common ratio

Given a₅ = 162 and a₈ = 4374 , then

a₁$r^{4}$ = 162 → (1)

a₁$r^{7}$ = 4374 → (2)

Divide (2) by (1)

$\frac{a_{1}r^{7} }{a_{1}r^{4} }$ = $\frac{4374}{162}$

r³ = 27

r = $\sqrt[3]{27}$ = 3

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

a₁$(3)^{4}$ = 162

81a₁ = 162

a₁ = $\frac{162}{81}$ = 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

$S_{n}$ = $\frac{a_{1}(r^{n}-1) }{r-1}$ , then

$S_{10}$ = $\frac{2(3^{10}-1) }{3-1}$

     = $\frac{2(59049-1)}{2}$

     = 59049 - 1

     = 59048