Question

The sum of 4 consecutive terms of an arithmetic progression is 32 and their product is 3465. Find the numbers.

Answer 1

Answer

The four numbers are 5, 7, 9, 11

Explanation

Let the four consecutive terms in A.P. be (a - 3d), (a - d), (a + d), (a + 3d)

Their common difference = 2d

first term = (a - 3d)

Given, the sum of those 4 terms is 32

⇒ (a - 3d) + (a - d) + (a + d) + (a + 3d) = 32

⇒ 4a = 32

⇒ a = 8

Now, their product is 3465

⇒ (a - 3d)(a - d)(a + d)(a + 3d) = 3465

⇒ (a - 3d)(a + 3d)(a - d)(a + d) = 3465

⇒ (a² - 9d²)(a² - d²) = 3465

(using the identity, (a + b)(a - b) = a² - b²)

⇒ a⁴ - a²d² - 9a²d² + 9d⁴ = 3465

⇒ a⁴ - 10a²d² + 9d⁴ = 3465

Now, put the value of a as 8

⇒ 8⁴ - 10(8)²d² + 9d⁴ = 3465

⇒ 4096 - 640d² + 9d⁴ = 3465

⇒ 9d⁴ - 640d² + 4096 - 3465 = 0

⇒ 9d⁴ - 640d² + 631 = 0

Now, let d²= y

Thus, we get

⇒ 9y² - 640y + 631 = 0

⇒ 9y² - 9y - 631y + 631 = 0 (splitting the middle term)

⇒ 9y(y - 1) - 631(y - 1) = 0

⇒ (y - 1)(9y - 631) = 0

⇒ y = 1 or y = 631/9

Now, put y as d²

⇒ d² = 1 or d² = 631/9

Since d² = 1 is a perfect square, we will take this value and neglect the other

⇒d = 1 or d = ( - 1)

Thus, the four numbers become (for d = 1)

a - 3d = 8 - 3 = 5

a - d = 8 - 1 = 7

a + d = 8 + 1 = 9

a + 3d = 8 + 3 = 11

For d = -1, we get the same numbers but in opposite order (11, 9, 7, 5)

Hence, the four numbers are 5, 9, 7 and 11