Determine the number of terms in the A.P. 3, 7, 11, ..., 399. Also, fi

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

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

Answer»

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.