a =?

Let the roots be z-d, z, z+d

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Sum of roots = $\frac{-(coefficient of x²)}{coefficient of x³}$

z-d+z+z+d=6a

z=2a

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Putting x=2a.

(2a)³-6a(2a)²+5a(2a)+88=0 (Since z is a root of the equation and z=2a)

8a³-24a³+10a²=-88

-16a³+10a²=-88

2a²(5-8a)=-88

a=2

As the three roots of the cubic equation are in a sequence the roots of the cubic equation

can be considered as r-d, r, r+d

So sum of roots = 3r = 6a

So r = 2a

The sum of roots taken two at a time = 5a can be written as

r (r-d) + r (r+d) + (r+d)(r-d) = r (2r) + r r - d d = 3 r r -d d

3 r r -d*d = 5a

So d d = 12 a a - 5 a

The product of the three roots is

(r r - d d)*r = -88

Rearranging all the three equations we get

   a*a(8a-5) = 44 = 4 * 11 and so a = 2;