Question

The sum of three consecutive numbers in AP is -3 and their product is 8 find the numbers​

Answer 1

Answer:

Solution :-

Let the number are x−y,x,x+y

Sum =−3

⇒x−y+3x+x+y=−3

⇒3x=−3

⇒x=−1

Now product =8

⇒(x−y)(x)(x+y)=8

Substituting x=−1

we get (−1−y)(−1)(−1+y)=8

(y

2

−1)=8

⇒y=±3

The no : are −4,−1,2 or 2,−1,−4

Answer 2

Given :

• Sum of three consecutive terms of AP = -3
• Product of three consecutive terms of AP = 8

To find :

Consecutive terms of AP

Note :

First of all we will assume terms of AP.

We can do this in 2 way

(1) Classical AP , which we have read - a , a+d , a+2d

(2) x-d , x , x+d

Now whenever, sum of consecutive terms are given , use (2) assumption .

SOLUTION :

So , Let , those three consecutive terms of AP are

x-d , x , x+d

Sum = (x-d) + x + (x+d)

➝ -3 = x + x + x - d + d

➝ -3 = 3x

➝ x = (-3)/3

➝ x = -1 equation 1

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Now , product = (x-d)(x)(x+d)

➝ 8 = (x)(x-d)(x+d)

[ Note - using identity (a-b)(a+b) = a² - b² ]

➝ 8 = x ( x² - d²)

[ Note - on putting value of x from equation 1 ]

➝ 8 = (-1) [ (-1)² - (d)² ]

➝ -8 = (1 - d² )

➝ - 8 - 1 = - d²

➝ d² = 9

➝ d = √9

➝ d = ±3

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Now ,

x = -1 and d = ±3

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When d = +3

• x - d = (-1) - (+3) = -1 - 3 = -4
• x = (-1)
• x + d = (-1) + (+3) = -1+3 = 2

So consecutive terms of AP are , -4 , -1 , 2

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When d = -3

• x - d = (-1) - (-3) = -1 + 3 = 2
• x = (-1)
• x + d = (-1) + (-3) = -1 - 3 = -4

So consecutive terms of AP are 2 , -1 , -4

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ANSWER :

Consecutive terms of AP are either of following

1. -4 , -1 , 2
2. 2 , -1 , -4