Algebra 2 geometric sequence arithmetic sequence?

Question

hree numbers form an arithmetic sequence whose constant differences 3. If the first number is increased by 1, the second increased by 6 , and the third by 19 the resulting 3 numbers form a geometric sequence.determine the 3 original numbers

Answer 1

Let the original sequence three numbers are a, a + 3, and a + 6.

If the first number is increased by 1, the second increased by 6 , and the third by 19

So, the geometric sequence three numbers are a + 1, a + 9, and a + 25.

Above numbers following geometric sequence so, t₂ / t₁ = t₃ / t₂

(a + 9)/(a + 1) = (a + 25)/(a + 9)

Cross multiplication.

(a + 9)(a + 9) = (a + 25)(a + 1)

FOIL method: the product of two binomials is the sum of the products of the First terms, the Outer terms, the Inner terms and the Last terms.

a² + 9a + 9a + 81 = a² + 1a + 25a + 25

a² + 18a + 81 = a² + 26a + 25

Subtract a² from each side.

18a + 81 = 26a + 25.

Subtract 18a from each side.

81 = 8a + 25

Subtract 25 from each side.

8a = 56

Divide each side by 8.

a = 7.

Therefore the original sequence is 7, 10, 13.

The geometric sequence is 8, 16, 32.