Answer :

Let first term be a and common difference be d.

We know that, a_{n} = a + (n - 1)d

∴ a_{6} = a + (6 - 1)d

⇒ 19 = a + 5d …(i)

and a_{17} = a + (17 - 1)d

⇒ 41 = a + 16d …(ii)

On solving eq. (i) and (ii), we get,

a = 9 and d = 2

⇒ a_{40} = a + (40 - 1)d

⇒ a_{40} = 9 + (40 - 1) × 2

⇒ a_{40} = 9 + 39× 2 = 9 + 78 = 87

Hence, 40^{th} term is 87.

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