Question

the sum of three consecutive terms in AP is 6 and their product is -125 find the three numbers​

Answer 1

Step-by-step explanation:

Let the three consecutive nos in AP be (a-d), a, (a+d).

Now, condition 1:

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

3a = 6

a = 6/3

a = 2

Next, condition 2:

(a-d) x a x (a+d) = - 125

$a \times ( {a}^{2} - {d}^{2} ) = - 125 \\ 2( {2}^{2} - {d}^{2} ) = - 125 \\ 4 - {d}^{2} = \: \frac{ - 125}{2} \\ 4 + \frac{125}{2} = {d}^{2} \\ \frac{133}{2} = {d}^{2} \: \\ d \: = \sqrt{ \frac{133}{2} } \\ a - d \: = 2 - \sqrt{ \frac{133}{2} } \: \\ \: \: \: \: \: \: \: \: \: \: \: \: \: = \sqrt{4} - \sqrt{ \frac{133}{2} } \\ = \frac{ \sqrt{8} - \sqrt{133} }{2} \\ a \: = 2 \\ a + d \: = 2 + \sqrt{ \frac{133}{2} } \: \\ \: \: \: \: \: \: \: \: \: \: \: \: \: = \sqrt{4} + \sqrt{ \frac{133}{2} } \\ = \frac{ \sqrt{8} + \sqrt{133} }{2} \\ hence \: the \: three \: consecutive \: \\ numbers \: in \: AP \: are \: \frac{ \sqrt{8} - \sqrt{133} }{2}, \\ \: 2 \: and \: \frac{ \sqrt{8} + \sqrt{133} }{2} \:$