Find the 6th term from end of the A.P. 17, 14, 11,… – 40.

Here, a = 17, d = 14 – 17 = –3 and l = –40

where l = a + (n – 1)d

Now, to find the 6^th term from the end, we will find the total number of terms in the AP.

So, –40 = 17 + (n – 1)(–3)

⇒ –40 = 17 –3n + 3

⇒ –40 = 20 – 3n

⇒ –60 = –3n

⇒ n = 20

So, there are 20 terms in the given AP.

Last term = 20^th

Second last term = 20 – 1 = 19^th

Third last term = 20 – 2 = 18^th

And so, on

So, the 6^th term from the end = 20 – 5 = 15^th term

So, a_n = a + (n – 1)d

⇒ a₁₅ = 17 + (15 – 1)(–3)

⇒ a₁₅ = 17 + 14 × –3

⇒ a₁₅ = 17 – 42

⇒ a₁₅ = –25