The sum Sn of the terms of an arithmetic progression is defined by the formula Sn = 4n2 – 3n for any n. Write the first three terms of the progression

Given: S_n = 4n² – 3n

Let first term = a and common difference = d

So S₁ = 4(1)² – 3(1)

= 4 -3

= 1

⇒ a = 1

S₂ = 4(2)² – 3(2)

⇒ a + a + d = 16 - 6

2a + d = 10

2(1) + d = 10

⇒ d = 8

So first three terms are, a, (a + d), (a + 2d)

=1, (1 + 8), (1 + 2(8))

= 1,9,17

The first three terms of the progression are 1,9,17