Question

find 3 consecutive term in an A.P whose sum is-3 and product of there cube is 27​

Answer 1

Answer:

Let the 3 consecutive terms are (a - d),a and (a + d).

The sum of 3 consecutive terms = -3

Therefore,

(a - d) + a + (a + d) = -3

a - d + a + a + d = -3

3a = -3

a = -3/3

a = -1 ...........(1)

The product of their cubes = 27

Therefore,

(a - d)'3 × a'3 × (a + d)'3 = 27

(a - d) × a × (a + d) = 3 .......(Taking cube root)

(-1 - d) × (-1) × (-1 + d) = 3 ......(From equation 1)

(-1 - d) × (-1 + d) = 3/-1

(-1 - d) × (-1 + d) = -3

(-1)'2 - d'2 = -3 ......[(a - b)(a + b) = a'2 - b'2]

1 - d'2 = -3

-d'2 = -3 -1

-d'2 = -4

d'2 = 4

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

Now,

The three consecutive terms are

(a - d) = -1 - 2 = -3

a = -1

(a + d) = -1 + 2 = 1

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