Determine the number of terms in the A.P. 3, 7, 11, ..., 399. Also, find its 20th term from the end.

Here, a = 3, d = 7 – 3 = 4 and l = 399

To find : n and 20^th term from the end

We have,

l = a + (n – 1)d

⇒ 399 = 3 + (n – 1) × 4

⇒ 399 – 3 = 4n – 4

⇒ 396 + 4 = 4n

⇒ 400 = 4n

⇒ n = 100

So, there are 100 terms in the given AP

Last term = 100^th

Second Last term = 100 – 1 = 99^th

Third last term = 100 – 2 = 98^th

And so, on

20^th term from the end = 100 – 19 = 81^st term

The 20^th term from the end will be the 81^st term.

So, t₈₁ = 3 + (81 – 1)(4)

t₈₁ = 3 + 80 × 4

t₈₁ = 3 + 320

t₈₁ = 323

Hence, the number of terms in the given AP is 100, and the 20^th term from the last is 323.