If a_n = 3 – 4n, then show that a₁, a₂, a₃, … from an AP. Also, find S₂₀.

Given that, nth term of the series is a_n = 3 - 4n

For a_1,

Put n = 1 so a₁ = 3 - 4(1) = - 1

For a_2,

Put n = 2, so a₁ = 3 - 4(2) = - 5

For a_1,

Put n = 3 so a₁ = 3 - 4(3) = - 9

For a_1,

Put n = 4 so a₁ = 3 - 4(4) = - 13

So AP is - 1, - 5, - 9, - 13, …

a₂ - a₁ = - 5 - (- 1) = - 4

a₃ - a₂ = - 9 - (- 5) = - 4

a₄ - a₃ = - 13 - (- 9) = - 4

Since, the each successive term of the series has the same difference. So, it forms an AP with common difference, d = - 4

We know that, sum of n terms of an AP is

Where a = first term

d = common difference

and n = no of terms

= 10[ - 2 - 76]

= - 780

So Sum of first 20 terms of this AP is - 780.