Find the 10th term form the end of the AP 4, 9, 14, .. , 254.

The above series of numbers forms an arithmetic progression with

first term(a) = 4 and,

common difference(d) = (n + 1)th term – nth term = 9 – 4 = 5

last term or nth term(a_n) = 254

Let the total no. of terms in above A.P be n.

∴ a_n = a + (n – 1) × d

⇒ 254 = 4 + (n – 1) × 5

⇒ 250 = 5n – 5

⇒ 5n = 255

∴ n = 51

∴ 10th term from the end of AP = 51 – 10 + 1 = 42th term from the beginning

∴ 42th term = a₄₂ = a + (42 – 1)d

= 4 + 41 × 5

= 209

Hence, 10th term from the end of AP is 209.