Question

the sum of three consecutive terms of AP is 6 AnD their product is -9. Find the terms

Answer 1

ANSWER

Step By Step Explanation

Let the 3 consecutive numbers be a-d, a, and a+d respectively.

where a = first term of the AP and d=is their common difference.

Now given that sun of the three numbers is 6.

So we get =>

a + a-d + a + d = 6

=> 3a = 6

=>a = 2

Also, a(a-d)(a+d) = -9

Putting a = 2 we get =>

2(2-d)(2+d) = -9

By using (a+b)(a-b) = a²-b² we get =>

2(2² - d²) = -9

=> 2(4- d²) = -9

=> 8 -2d² = -17

=>-2d² = -25

=>d² = 25/2

=> d =√(25/2)

Hence the terms are =>

2 - √(25/2),2 and 2 + √(25/2) respectively.

Also we can write the terms as

2 - (5/√2) , 2 and 2 + (5/√2)

Remember

1)An = a +(n-1)d

where An = nth term of AP.

2)Sn =$\dfrac{n}{2}$[2a + (n-1)d].

where Sn is the sum of n terms of AP.

Answer 2

Answer:

Step By Step Explanation

Let the 3 consecutive numbers be a-d, a, and a+d respectively.

where a = first term of the AP and d=is their common difference.

Now given that sun of the three numbers is 6.

So we get =>

a + a-d + a + d = 6

=> 3a = 6

=>a = 2

Also, a(a-d)(a+d) = -9

Putting a = 2 we get =>

2(2-d)(2+d) = -9

By using (a+b)(a-b) = a²-b² we get =>

2(2² - d²) = -9

=> 2(4- d²) = -9

=> 8 -2d² = -17

=>-2d² = -25

=>d² = 25/2

=> d =√(25/2)

Hence the terms are =>

2 - √(25/2),2 and 2 + √(25/2) respectively.

Also we can write the terms as

2 - (5/√2) , 2 and 2 + (5/√2)

Remember

1)An = a +(n-1)d

where An = nth term of AP.

2)Sn =[2a + (n-1)d].

where Sn is the sum of n terms of AP.

Hope it help you