Question

Three numbers are in AP there sum is 48 and product of 1st and 3rd term is 175.find the number

Answer 1

let the first 3 terms of the A.P be a-d, a and a+d

by data, (a-d) + a + (a+d)= 48

3a=48

a=16

as per data: (a-d)*a= 4(a+d)+12--(1)

substituting the value of 'a' in (1); (16-d)*16=4(16+d) +12

256 - 16d = 64 + 4d + 12

256 - 16d =76 + 4d

256-76= 16d+4d

180 = 20d

d= 9

Answer 2

Answer:

$\large{\underline{\boxed{\bf (7,16,25) \: or \: (25,16,7)}}}$

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 = 48

⇒ a = $\sf \dfrac{48}{3}$

⇒ a = 16

Also, the product of first and third term is 175.

⇒ (a - d) * (a + d) = 175

⇒ (a² - d²) = 175

⇒ (16² - d²) = 175

⇒ 256 - d² = 175

⇒ d² = 256 - 175

⇒ d² = 81

⇒ d = ± √81

⇒ d = ± 9

Thus, a = 16 and d = ± 9.

The required numbers are;

(a - d) = (16 - 9) = 7

a = 16

(a + d) = 16 + 9 = 25

Or, the other possibility can be -

(a - d) = 16 + 9 = 25

a = 16

(a + d) = 16 - 9 = 11

$\rule{300}{2}$

Hence, the required numbers in an AP are (7, 16, 25) or (25, 16, 7).