Answer 1

Given S_n = 2n² – 3n

S₁ = 2(1)² – 3(1) = 2 – 3 = – 1

⇒ t₁ = a = – 1

S₂ = 2(2²) – 3(2) = 8 – 6 = 2

t₂ = S₂ – S₁ = 2 – (– 1) = 3

∴ d = t₂ – t₁ = 3 – (– 1) = 4

Consider a, a + d, a + 2d, ….….

– 1, – 1 + 4, – 1 + 2(4), …..…

– 1, 3, 7, ….

Clearly, this is an A.P. with a = – 1, and d = 4.