Question

the sum of three numbers in AP is -3 and their product is 8. find those numbers.

Answer 1

Let the first term be a-d.

Let the difference be d.

So AP=a-d,a,a+d...................

Sum of the first three is -3

So:

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

3a=-3

a=-3/3

=-1

Now the product is 8.

(a-d)(a)(a+d)=8

But wait a=-1

So:

(-1-d)(-1)(-1+d)=8

=-1(1+d)(-1)(d-1)=8

-1*-1=1

so:

=(d+1)(d-1)=8

=d²-1=8

=d²=9

=d=+3 or d=-3

For d=3

AP=a-d,a,a+d

a=-1

So

AP=-1-3,-1,-1+3

=-4,-1,2

Also when d=-3

AP=-1+3,-1,-1-3

=2,-1,-4

Thus these are the APs of the given problem.

(-4,-1,2) or (2,-1,-4)

Hope it helps.

Answer 2

$\tt{Here \ is \ the \ answer}$

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$<b>$
Let the numbers be (a + 3d) , a and
(a - 3d).

A/q

Sum of these numbers = -3 (given)

So,

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

3a = -3

a = -1 ______ eq (1)

Also,

Product of these numbers = 8

So,

(a + 3d)(a)(a - 3d) = 8

(a + 3d)(a - 3d)(a) = 8

[(-1) + 3d] [(-1) - 3d] (-1) = 8 _______ From eq (1)

=> Using identity -

=> (x + y)(x - y) = x² - y²

So,

(-1)² - (3d)² (-1) = 8

(1 - 9d²)(-1) = 8

-1 + 9d² = 8

9d² = 9

d² = 1

d = +1 or -1

Case I :-

When a = -1 and d = 1

a + 3d = -1 + 3(1) = -1 + 3 = 2
a = -1
a - 3d = -1 - 3(1) = -1 - 3 = -4

Case II :-

When a = -1 and d = -1

a + 3d = -1 + 3(-1) = -1 - 3 = -4
a = -1
a - 3d = -1 - 3(-1) = -1 + 3 = 2

Hence, the numbers are -1, 2 and -4.

$\rule{150}{5}$

$\tt{Hope \ this \ helps.}$