Question

P q r are three consecutive integers and averave of p q r is 3

Answer 1

let p = x, q=x+1, r=x+2

average of the pqr = 3

(x+x+1+x+2)/3=3

$\frac{x + x + 1 + x + 2}{3} = 3 \\ \frac{3x + 3}{3} = 3 \\ \frac{3(x + 1)}{3} = 3 \\ x + 1 = 3 \\ x = 3 - 1 \\ x = 2$

p = 2, q = 3, r = 4

Answer 2

Answer:

Value of p is 2, value of q is 3 and value of r is 4.

Step-by-step explanation:

Let, p equal to x, q equal to (x + 1) and  r equal to (x + 2) because p, q and r are three consecutive integers

Average of p, q, and r is [x + (x+1) + (x+2)] / 3

And according to the question, average of p, q, and r is 3

So, [x + (x+1) + (x+2)] / 3 = 3

⇒ (3x + 3) / 3 = 3

⇒ 3 (x + 1) /3 = 3

⇒ (x + 1) = 3

⇒ x = 3 - 1

⇒ x = 2

So, p = 2,  q = 2+1 = 3,  r = 2+2 = 4