Question

Please help me with this.​

Answer 1

Step-by-step explanation:

We have,

$\tt{ \blue{ {x}^{3} - 12 {x}^{2} + 3x + 10 }}$

Its zeroes are $\sf{\green{(a-d), a, (a+d)}}$

Now,

$\sf{ \color{orange} Sum \: \: \: of \: \: \: zeros \: , (a - d) + a + (a + d) = \dfrac{ - ( - 12)}{1} }$

$\sf{ \implies (a - d) + a + (a + d) = 12 }$

$\sf{ \implies a - d + a + a + d= 12 }$

$\sf{ \implies 3a = 12 }$

$\sf{ \implies a = 4 }$