Question

8) Given x+8, x+3, x+1 are three consecutive terms of a geometric sequence. Determine the value of x and the 6th term.​

Answer 1

x =
`(1)/(3)` , a₆ =
`(32)/(375)`

Explanation:

the nth term of a geometric sequence is

`a_(n)` = a₁
`r^(n-1)`

where a₁ is the first term and r the common ratio

r =
`(a_(2) )/(a_(1) )` =
`(a_(3) )/(a_(2) )` = ... =
`(a_(n+1) )/(a_(n) )`

here a₁ = x + 8 , a₂ = x + 3 and a₃ = x + 1 , then

r =
`(x+3)/(x+8)` =
`(x+1)/(x+3)` ( cross- multiply )

(x + 3)² = (x + 1)(x + 8) ← expand factors using FOIL

x² + 6x + 9 = x² + 9x + 8 ( subtract x² + 9x from both sides )

- 3x + 9 = 8 ( subtract 9 from both sides )

- 3x = - 1 ( divide both sides by - 3 )

x =
`(1)/(3)`

then r =
`(a_(2) )/(a_(1) )` =
`((1)/(3)+3 )/((1)/(3)+8 )` ( multiply numerator/ denominator by 3 to clear the fractions )

r =
`(1+9)/(1+24)` =
`(10)/(25)` =
`(2)/(5)`

and

a₁ = x + 8 =
`(1)/(3)` + 8 = 8
`(1)/(3)` =
`(25)/(3)`

Then

a₆ =
`(25)/(3)` ×
`((2)/(5)) ^(5)` =
`(25)/(3)` ×
`(32)/(3125)` ( cancel 25 and 3125 by 25 )

a₆ =
`(1)/(3)` ×
`(32)/(125)` =
`(32)/(3(125))` =
`(32)/(375)`