Given,

6^th term of an A.P is 19 and 17^th terms of an A.P. is 41

⇒ a₆ = 19 and a₁₇ = 41

We know, a_n = a + (n – 1)d where a is first term or a₁ and d is common difference and n is any natural number

When n = 6:

∴ a₆ = a + (6 – 1)d

⇒ a₆ = a + 5d

Similarly, When n = 17:

∴ a₁₇ = a + (17 – 1)d

⇒ a₁₇ = a + 16d

According to question:

a₆ = 19 and a₁₇ = 41

⇒ a + 5d = 19 ………………(i)

And a + 16d = 41…………..(ii)

Subtracting equation (i) from (ii):

a + 16d – (a + 5d) = 41 – 19

⇒ a + 16d – a – 5d = 22

⇒ 11d = 22

⇒ d = 2

Put the value of d in equation (i):

a + 5(2) = 19

⇒ a + 10 = 19

⇒ a = 19 – 10

⇒ a = 9

As, a_n = a + (n – 1)d

a₄₀ = a + (40 – 1)d

⇒ a₄₀ = a + 39d

Now put the value of a = 9 and d = 2 in a₄₀

⇒ a₄₀ = 9 + 39(2)

⇒ a₄₀ = 9 + 78

⇒ a₄₀ = 87

Hence, 40^th term of the given A.P. is 87