Question

The sum of three consecutive terms in an A.P is 6 and their product is -120. Find the three numbers.

Answer 1

a1= (a-d) = -6

a2= (a) =2

a3= (a+d) = 10

Step-by-step explanation:

a1+a2+a3=6 ...(1)

a1xa2xa3= -120 ...(2)

therefore, let's assume (a-d),a,(a+d) to be a1, a2, a3 respectively,

therefore,

in (1),

a-d+a+a+d=6

3a=6

a=2

therefore,

in (2),

(a-d)(a+d)(a)= -120

(a²-d²)(a) = -120

therefore, substituting value of a,

therefore,

(2²-d²)2= -120

4-d²= -60

d²=64

therefore,

d=8

therefore,

(a)=2

(a)=2(a-d)= -6

(a)=2(a-d)= -6(a+d)= 10

I hope that helps!!

Answer 2

Given :

• Sum of three consecutive terms in an AP is 6
• And their product is -120

To find :

The three numbers

Solution:

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

where,

a = first term of the AP

d=is their common difference

According to the question:

Sum of consecutive terms in an AP is 6

$\implies\sf\:a+(a+d)+(a-d)=6$

$\implies\sf\:3a=6$

$\implies\sf\:a=\dfrac{6}{3}=2...(1)$

And their product is -120

$\implies\sf\:(a)\times(a-d)\times(a+d)=-120$

we know that (a+b)(a-b)= a²-b²

$\implies\tt\:a(a^2-d^2)=-120$

Now put a=2 from equation (1)

$\implies\tt\:2(4-d^2)=-120$

$\implies\tt8-2d^2=-120$

$\implies\tt128=2d^2$

$\implies\tt\:d^2=\dfrac{128}{2}$

$\implies\tt\:d=\sqrt{64}=\pm8$

Case 1: if d =8

Then ,

a-d = 2-8=-6 ,a=2 and a+d=2+8= 10

Hence ,terms are -6,2,10

Case 2: if d=-8

Then ,

a-d =2-(-8)=10, a=2 and a+d = 2-8=-6

Hence ,terms are 10,2,-6