The 10th and 18th terms of an A.P are 41 and 73 respectively. Find the 27th term.

Here, t₁₀ = 41 and t₁₈ = 73

We know that nth term of A.P. , t_n = a + (n – 1) d.

First, t₁₀ = a + (10 – 1) d = 41

⇒ a + 9d = 41 … (1)

Then, t₁₈ = a + (18 – 1) d = 73

⇒ a + 17d = 73 … (2)

From (1) and (2),

a + 9d = 41

a + 17d = 73

(-) (-) (-)

-8d = -32

⇒ d = 32/8 = 4

Substituting d = 4 in (1),

⇒ a + 9(4) = 41

⇒ a = 41 – 36 = 5

Now, t₂₇ = 5 + (27 – 1) (4)

= 5 + 26(4)

= 5 + 104

= 109

∴ The 27^th term i.e. t₂₇ = 109