Given: a₂ = a + d = ….. (1)

a₆ = a + d = - ….. (2)

Solving equations (2) from (1) we get

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Now substituting, d = – in equation (1), we get

a + (- ) = ⇒ a = + =

Thus,

a₁ = a = ;

a₃ = a + 2d = + × (- ) = – = ;

a₄ = a + 3d = + × (- ) = – = ;

a₅ = a + 4d = + × (- ) = – = –