Question

the sum of three numbers in A.P is -3 and their product is 8. find the numbers

Answer 1

Let the three numbers in AP are  a-d, a, a+d where d is the common difference
Their sum = -3
a-d+a+a+d = -3
3a = -3
a = -1
their product = 8
(a+d)(a)(a+d) = 8

a(a²-d²) = 8
(-1)(1-d²) = 8
1-d² = -8
d² = 9
d = +3 or -3

Numbers are when d = -3
(-1+3), -1, (-1-3)  = 2, -1, -4
Numbers are when d = 3
(-1-3), -1, (-1+3)  = -4, -1, 2

Answer 2

The series is -4, -1, +2

Given:

Sum of three numbers in “AP” is -3

The product of three numbers in “AP” is 8

To find:

Calculate the three numbers which is used in that AP

Answer:

Let the numbers are

(a - d), a, (a + d)

$Sum = (a - d) + a + (a + d)$

$3 \times a= -3$

$a = - 1$

$Product= (a - d) \times (a) \times (a + d)$

$(a^{2} - a \times d) \times (a + d) = 8$

$a^{3} + a^{2}\times d - a^{2} \times - a \times d^{2} = 8$

$a^{3} - a \times d^{2} = 8$

Substitute, $a = - 1$

$(-1)^{3} - (-1) \times d^{2} = 8$

$-1 + d^{2} = 8$

$d^{2} = 9$

$d = 3$

Therefore the numbers are,

$(a - d) = (-1 -3) = -4$

$a = -1$

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