Given a and d for the following A.P., find the following A.P.:

a = –3, d = –2

Formula Used.

a_n = a + (n–1)d

If a = –3, d = –2

Then a₁ = a + (n–1)d = –3 + (1–1)(–2) = –3

a₂ = a + (n–1)d = –3 + (2–1)(–2) = –3 + 1×(–2) = –5

a₃ = a + (n–1)d = –3 + (3–1)(–2) = –3 + 2×(–2) = –7

a₄ = a + (n–1)d = –3 + (4–1)(–2) = –3 + 3×(–2) = –9

a_n = a + (n–1)d = –3 + (n–1)(–2) = –1–2n

∴ The A.P is –3, –5, –7, –9……, –1–2n