The sum of the second and third terms of a geometric progression is six times the fourth term. 1. FIND THE POSSIBLE VALUES OF THE COMMON RATIO.
2. IF THE SECOND TERM IS 8 AND THE COMMON RATIO IS POSITIVE, FIND THE FIRST SEVEN TERMS.

Step-by-step explanation:

a2 = a1×r

a3 = a1×r²

a1×r + a1×r² = 6×a4 = 6×a1×r³

1.

r + r² = 6r³

6r³ - r² - r = 0

r×(6r² - r - 1) = 0

the first solution is obvious : r = 0.

but this is no useful ratio for a geometric sequence.

the other 2 solutions are in

6r² - r - 1 = 0

the general solutions for a quadratic equation are

(-b ± sqrt(b² - 4ac))/(2a)

in our case

a = 6

b = -1

c = -1

so,

(1 ± sqrt(1 - 4×6×-1))/12

r = (1 ± sqrt(25))/12

r = (1 ± 5)/12

r1 = (1+5)/12 = 6/12 = 1/2

r2 = (1-5)/12 = -4/12 = -1/3

2.

we can ignore r2 (negative) and just focus on r1 (1/2).

the second term is 8. that means

a2 = 8 = a1×r = a1×1/2

a1 = 2×a2 = 16

so, we have

a1 = 16

a2 = 8

a3 = a2×1/2 = 8×1/2 = 4

a4 = a3×1/2 = 4×1/2 = 2

a5 = a4×1/2 = 2×1/2 = 1

a6 = a5×1/2 = 1×1/2 = 1/2

a7 = a6×1/2 = 1/2 × 1/2 = 1/4

the polynomial is a perfect square trinomial of the form a^2-2ab+b^2:

option a. true

step-by-step explanation:

9x^2-12x+4

comparing with the form a^2-2ab+b^2

1.) a^2=9x^2

solving for a: square root both sides of the equation:

sqrt(a^2)=sqrt(9x^2)

a=sqrt(9) sqrt(x^2)

a=3x

2.) b^2=4

solving for b: square root both sides of the equation:

sqrt(b^2)=sqrt(4)

b=2

to be a perfet square trinomial 2ab must be equal to 12x. let's check:

2ab = 2 (3x) (2)

2ab = 12x

correct

the solution is x=1

step-by-step explanation:

we are given our   equation as

let's assume

left side function is

right side function is

now, we can draw graph

and then we can get intersection point

the intersection point will be solution

so, the solution is

x=1