Since, nth term is given as (5 - 6n)

Put n = 1, we get a₁ = - 1 = first term

Put n = 2, we get a₂ = - 7 = second term

Now, d = a₂ - a₁ = - 7 - (-1) = - 6

Sum of first n terms = S_n= [2a + (n - 1)d]; where a is the first term

and d is the common difference.

= [ - 2 + (n - 1)(-6)]

= n[ - 1 - 3n + 3]

= n(2 - 3n)

∴ sum of first 20 terms = S₂₀

= [2(-1) + (20 - 1)(-6)]

= 10 × [ - 2 - 114]

= 10 × [ - 116]

= - 1160