Question

three numbers are in geometric sequence. the difference between the third number and the first number is 45 and the product of these two is 2916. find the numbers

Answer 1

ANSWER:

Given:

• 3 numbers are in GP
• Difference of 3rd and 1st terms is 45.
• Product of 1st and 3rd term is 2196.

To Find:

• The numbers

Solution:

Let the numbers be a/r, a and ar.

We are given that,

⇒ (ar) × (a/r) = 2916

⇒ a² = 2916

Taking square root,

⇒ a = √(2916)

⇒ a = 54

We are also given that,

⇒ (ar) - (a/r) = 45

Taking a = 54,

⇒ 54r - 54/r = 45

⇒ (54r² - 54)/r = 45

⇒ 54r² - 54 = 45r

⇒ 6r² - 6 = 5r

⇒ 6r² - 5r - 6 = 0

⇒ 6r² - 9r + 4r - 6 = 0

⇒ 3r(2r - 3) + 2(2r - 3) = 0

⇒ (2r - 3)(3r + 2) = 0

⇒ 2r - 3 = 0  &  3r + 2 = 0

⇒ 2r = 3  &  3r = -2

⇒ r = 3/2  &  r = -2/3

So, the terms are:

1. For r = 3/2

• a/r = 54/(3/2) = 54 * 2/3 = 36
• a = 54
• ar = 54 * 3/2 = 81

2. For r = -2/3

• a/r = 54/(-2/3) = 54 * -3/2 = -81
• a = 54
• ar = 54 * -2/3 = -36

Therefore, the numbers are 36, 54 and 81 or -81, 54 and -36.