# 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 20th 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 = 100th

Second Last term = 100 – 1 = 99th

Third last term = 100 – 2 = 98th

And so, on

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

The 20th term from the end will be the 81st 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 20th term from the last is 323.

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