Question

The sum of three numbers of a ap of -6 and their product is 64 . Find those numbers

Answer 1

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Answer 2

Answer:

Numbers are  4, -2, -8.

Step-by-step explanation:

Since the given three terms are in AP.

'a' be the first term and 'd' be the common difference then,

Let numbers be (a-d) , a , (a+d)

Given sum of three numbers be -6

Then , (a-d) + a + (a+d) = -6

3a = -6

⇒ a = - 2

thus, the numbers becomes  (-2-d) , (-2) , (-2+d)

Also given product of terms is 64.

$(-2-d) \times (-2) \times (-2+d) = 64$

$(-2-d) \times (-2+d) = -32$

Using identity,  $(a+b)(a-b)=a^2-b^2$

Here, a = -2 and b = d

$(-2)^2-d^2=-32$

$4-d^2=-32$

$-d^2=-32-4$

$-d^2=-36$

$d^2=36$

$d=\sqrt{36}$

d = ±6

Thus, numbers are ,

Case 1: a = -2 , d = 6

(a-d) , a , (a+d) = -8, -2, 4

Case 2: a = -2 , d= - 6

(a-d) , a , (a+d) = 4, -2, -8

Thus, Numbers are  4, -2, -8.