Question

If sum of the three numbers in an ap is 21 and the product is 231.Find the numbers.

Answer 1

Hello user

Let the three numbers be a-d, a, a+d

A.T.Q , we have

3a = 21
a = 7

Also,

(7-d)(7)(7+d) = 231

49 - d^2 = 33

d^2 = 16

d = +4 and - 4

So, the numbers will be either

3,7 and 11 or 11 , 7 and 3

Hoope it works

Answer 2

Answer:

$\large{\underline{\boxed{\bf (3,7,11) \: or \: (11,7,3)}}}$

Step-by-step explanation:

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

It is given that ;

Sum of numbers is 21.

⇒ a - d + a + a + d = 21

⇒ 3a = 21

⇒ a = $\sf \dfrac{21}{3}$

⇒ a = 7

Also, the product of numbers is 231.

⇒ (a - d) * a * (a + d) = 231

⇒ a (a² - d²) = 231

⇒ 7 (7² - d²) = 231

⇒ (49 - d²) = $\sf \dfrac{231}{7}$

⇒ 49 - d² = 33

⇒ d² = 49 - 33

⇒ d² = 16

⇒ d = ± √16

⇒ d = ± 4

Thus, a = 7 and d = ± 4.

The required numbers are;

• (a - d) = (7 - 4) = 3
• a = 7
• (a + d) = 7 + 4 = 11

Or, the other possibility can be -

• (a - d) = 7 + 4 = 11
• a = 7
• (a + d) = 7 - 4 = 3

$\rule{300}{2}$

Hence, the required numbers in an AP are (3, 7, 11) or (11, 7, 3).