Which term of the A.P. 3, 14, 25, 36, …., will be 99 more than its 25^th term?

In the given AP, the first term = a = 3

Common difference = d = 14 – 3 = 11

To find: place of the term which is 99 more than its 25^th term.

So, we first find its 25^th term.

Since, we know that

a_n = a + (n-1) × d

∴ a₂₅ = 3 + (25 -1) × 11

⇒ a₂₅ = 3 + 24 × 11

⇒ a₂₅ = 3 + 264

⇒ a₂₅ = 267

∴ 25^th term of the AP is 267.

Now, 99 more than 25^th term of the AP is 99 + 267 = 366.

So, to find: place of the term 366.

So, let a_n = 366

Since, we know that

a_n = a + (n-1) × d

∴ 366 = 3 + (n -1) × 11

⇒ 366 – 3 = 11n - 11

⇒ 363 = 11n - 11

⇒ 363 + 11 = 11n

⇒ 11n = 374

⇒ n = 374/11 = 34

∴ 34^th term of the AP is the term which is 99 more than 25^th term.