Question

The sum of the second and third terms of a geometric progression is six times the fourth term.

Answer 1

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