Given: 10^th term of an A.P is 41, and 18^th terms of an A.P. is 73

⇒ a₁₀ = 41 and a₁₈ = 73

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

When n = 10:

∴ a₁₀ = a + (10 – 1)d

⇒ a₁₀ = a + 9d

Similarly, When n = 18:

∴ a₁₈ = a + (18 – 1)d

⇒ a₁₈ = a + 17d

According to question:

a₁₀ = 41 and a₁₈ = 73

⇒ a + 9d = 41 ………………(i)

And a + 17d = 73…………..(ii)

Subtracting equation (i) from (ii):

a + 17d – (a + 9d) = 73 – 41

⇒ a + 17d – a – 9d = 32

⇒ 8d = 32

⇒ d = 4

Put the value of d in equation (i):

a + 9(4) = 41

⇒ a + 36 = 41

⇒ a = 41 – 36

⇒ a = 5

As, a_n = a + (n – 1)d

a₂₆ = a + (26 – 1)d

⇒ a₂₆ = a + 25d

Now put the value of a = 5 and d = 4 in a₂₆

⇒ a₂₆ = 5 + 25(4)

⇒ a₂₆ = 5 + 100

⇒ a₂₆ = 105

Hence, 26^th term of the given A.P. is 105