Question

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

Answer 1

Answer:

Three consecutive number are -6, 2 ,10

Step-by-step explanation:

Let the three consecutive number be (a-d), a ,(a+d)

According to first given condition

a+d+a+a-d =6

3a=6

a=2 ..................(1)

put the values in the assumed consecutive number then it will be

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

According to second condition

(2-d)×2×(2+d)= -120

(4-d)(2-d)= -120

8+4d-4d-2d^2= -120

-2d^2= -120-8

d^2=-128/-2

d^2=64

d = 8..............(2)

from (1) and (2)

The three consecutive number are

a-d =2-8 =-6

a=2

a+d = 2+8 = 10

Answer 2

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GIVEN :

The sum of three terms of an AP = 6

The product of them = - 120.

TO FIND :

The terms.

SOLUTION:

Let the three terms be (a - d) , (a ), (a + d ).

Where a is the first term of an AP

d is the common difference of an AP.

=> a - d + a + a + d = 6

3a = 6

a = 6/3

a = 2

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

a^3 - a (d^2) = - 120

(2) ^3 - 2( d^2) = - 120

8 + 120 = 2 (d^2)

128 = 2 (d^2)

64 = d^2

d = 8

The three terms are ,

=> 2 - 8 , 2 , 2 + 8

=> -6 , 2 , 10

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