Question

Find the three numbers in



a.P whose sum is 12 and product is 48

Answer 1

Answer:

$\large{\underline{\boxed{\bf (2,4,6) \: or \: (6,4,2)}}}$

Step-by-step explanation:

Let the required numbers be (a - d), a and (a + d).

It is given that ;

Sum of numbers is 12

⇒ a - d + a + a + d = 12

⇒ 3a = 12

⇒ a = $\sf \dfrac{12}{3}$

⇒ a = 4

Also, the product of numbers is 48.

⇒ (a - d) * a * (a + d) = 48.

⇒ a (a² - d²) = 48

⇒ 4 (4² - d²) = 48

⇒ (16 - d²) = $\sf \dfrac{48}{4}$

⇒ 16 - d² = 12

⇒ d² = 16 - 12

⇒ d² = 4

⇒ d = ± √4

⇒ d = ± 2

Thus, a = 4 and d = ± 2.

The required numbers are;

(a - d) = (4 - 2) = 2

a = 4

(a + d) = 4 + 2 = 6

Or, the other possibility can be -

(a - d) = 4 + 2 = 6

a = 4

(a + d) = 4 - 2 = 2

$\rule{300}{2}$

Hence, the required numbers in an AP are (2, 4, 6) or (6, 4, 2).

Answer 2

Refer the attachment.