For A.P., S_n — 2S_{n – 1} + S_{n – 2} =…. (n > 2)

We can recall that, S_n = × (2a + (n – 1)d) ……(1)

So now, S_{n – 1} = × (2a + ((n – 1) – 1)d)

⇒ S_{n – 1} = × (2a + (n – 2)d) …….(2)

So now, S_{n – 2} = × (2a + ((n – 2) – 1)d)

⇒ S_{n – 2} = × (2a + (n – 3)d) ………(3)

Now putting the above values in the equation_,

We get,

⇒ S_n — 2S_{n – 1} + S_{n – 2}

= × (2a + (n – 1)d) – 2[ × (2a + (n – 2)d)] + × (2a + (n – 3)d)

= – 2[] +

= an + – – 2an – dn² + 2a – 2d + 3nd + an + – – 2a + 3d

= [an – 2an + an] + [ – dn² + ] + [2a – 2a] + [ – + 3nd – ] + [ – 2d + 3d]

= 0 + 0 + 0 + 0 + d

= d

∴ S_n — 2S_{n – 1} + S_{n – 2} = d

∴ the correct option is (b)