Question

three number in an AP the sum of three number is 27 and product 648 find the number​

Answer 1

AnswEr :

The required numbers are 6 , 9 and 12

SoluTion :

Let us consider the first term be a and common difference be d

Therefore , the three numbers will be : a - d , a and a + d

According to question

$\sf \implies a - d + a + a + d = 27 \\\\ \sf \implies 3a = 27 \\\\ \sf \implies a = 9$

Again , the product is 648

$\sf \implies (a - d)a(a + d) = 648 \\\\ \sf \implies a(a^{2} - d^{2} ) = 648 \\\\ \sf \implies 9(9^{2} - d^{2} ) = 648 \\\\ \sf \implies 81- d^{2} = 72 \\\\ \sf \implies -d^{2} = 72 - 81\\\\ \sf \implies d^{2} = 9 \\\\ \sf \implies d = 3$

Thus , common difference, d is 3

Therefore , the required numbers are :

9 - 3 , 9 and 9 + 3

→ 6 , 9 and 12

Answer 2

Answer:

three numbers are :

x-d = 9-3 = 6

x = 9

x+d = 9+3 = 12

Step-by-step explanation:

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

Sum of three numbers = 27

product of three numbers = 648

acc to the question

x-d + x + x+d = 27

3x = 27

x= 27/3

x = 9

Now,

(x-d) * x * (x+d) = 648

(x^2-xd) * (x+d)= 648

Putting the value of x

(81 - 9d) * (9 + d) = 648

729 - 81d + 81d - 9d^2 = 648

729 - 9d^2 = 648

-9d^2 = - 81

9d^2 = 81

d^2 = 9

d = 3

So,

three numbers are :

x-d = 9-3 = 6

x = 9

x+d = 9+3 = 12