The arithmetic mean of the first and third members of some geometric progression is 4 more

The arithmetic mean of the first and third members of some geometric progression is 4 more than the second term of this progression. The difference between the second and the first slice of the progression is 4. Find the sixth term of the progression.

Let us denote the first term of this geometric progression by the letter b, and the denominator of the progression by the letter q.

Then the second term of the progression will have the form: b2 = b * q, and the third, respectively, b3 = b * q²

By the condition of the problem, the difference between the second and first terms is 4, which means:

b * q – b = 4,

b * (q – 1) = 4,

b = 4 / (q – 1), q is not equal to 1.

The arithmetic mean of the first and third terms is 4 more than the second term of the progression:

(b + b * q²) / 2 = b * q + 4,

b + b * q² – 2 * b * q = 8,

b * q² + b * (1 – 2 * q) = 8.

Substitute the expression b into this equation:

4 * q² / (q – 1) + 4 * (1 – 2 * q) / (q – 1) = 8,

4 * q² + 4 – 8 * q = 8 * (q – 1),

4 * q² – 16 * q + 12 = 0,

q² – 4 * q + 3 = 0.

Let’s find the discriminant: D = 4² – 4 * 1 * 3 = 16 – 12 = 4, so

q = (4 + 2) / 2 = 3 and q = (4 – 2) / 2 = 1.

Since q cannot be equal to 1, the equation has one solution q = 3.

We can now find b:

b = 4 / (3 – 1) = 2.

The sixth term of the progression will be equal to:

b6 = 2 * 3⁵ = 2 * 243 = 486.

Answer: b6 = 486.