Let the zeros of the given polynomial be α, β and γ. (3 zeros as it’s a cubic polynomial) 

And given, the zeros are in A.P. 

So, let’s consider the roots as 

α = a – d, β = a and γ = a + d 

Where, a is the first term and d is the common difference. 

From given f(x), a = 2, b = -15, c = 37 and d = 30 

⇒ Sum of roots = α + β + γ 

= (a – d) + a + (a + d) 

= 3a 

= (\(\frac{-b}{a}\)) 

= -(\(\frac{-15}{2}\)) 

= \(\frac{15}{2}\) 

So, calculating for a, we get 3a = \(\frac{15}{2}\) 

⇒ a = \(\frac{5}{2}\) 

⇒ Product of roots = (a – d) x (a) x (a + d) 

= a(a² –d²) 

= \(\frac{-d}{a}\)

= \(\frac{-(30)}{2}\) 

= 15 

⇒ a(a² –d²) = 15 

Substituting the value of a, we get 

⇒ (\(\frac{5}{2}\))[(\(\frac{5}{2}\))² –d²] = 15 

⇒ 5[(\(\frac{25}{4}\)) –d²] = 30 

⇒ (\(\frac{25}{4}\)) – d² = 6 

⇒ 25 – 4d² = 24 

⇒ 1 = 4d² 

∴ d = \(\frac{1}{2}\) or -\(\frac{1}{2}\)

Taking d = \(\frac{1}{2}\) and a = \(\frac{5}{2}\) 

We get, 

the zeros as 2, \(\frac{5}{2}\) and 3 

Taking d = -\(\frac{1}{2}\) and a = \(\frac{5}{2}\) 

We get, the zeros as 3, \(\frac{5}{2}\) and 2.