Question

the sum of three numbers in ap is 12and their product is 48.find the numbers

Answer 1

let the three numbers be :

a+d , a , a-d

given the sum is 12 so add the numbers

3a = 12

a = 4 ( therefore the first number of the ap is 4)

(a+d)(a)(a-d) = 48 ( we take a to the left hand side it becomes division)

(a2 - d2) = 48/4

(42 - d2) = 12

16 - d2 = 12

-d2 = -4

d= 2

therefore the numbers are 2,4,6.

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Answer 2

Answer:

Step-by-step explanation:

I hope the question is

the sum of three consecutive numbers in ap is 12and their product is 48.find the numb=ders.

let x be the middle number and d be the common difference

then the numbers are x-d, x and x+d

given their sum = 12

sum of three consecutive terms in an ap is same as three times the middle number.

so 3x=12

⇒ x=4

their product = 48

⇒(x-d)x(x+d)=48

⇒x(x²-d²)=48

⇒4(16-d²)=48

⇒4d²=64-48=16

⇒d²=4

⇒d=±2

so the numbers are 2, 4, 6