Find the eleventh term from the last term of the AP:

27, 23, 19, ..., –65.

We have the AP: 27, 23, 29, …, -65

Here, first term of the series (a) = 27

Common difference of the series (d) = 23 – 27 = -4

Last term of the series (a_n) = -65

We need to find position of -65, i.e., n.

We know the formula,

a_n = a + (n – 1)d

⇒ -65 = 27 + (n – 1)(-4)

⇒ -65 = 27 – 4n + 4

⇒ 4n = 65 + 27 + 4

⇒ 4n = 96

⇒ n = 96/4 = 24

∴ -65 is in the 24^th position.

To find eleventh term from the last term,

Position of last term = 24^th

Position of second last term = (24 – 1)^th = 23^rd

Position of third last term = (24 – 2)^th = 22^nd

Similarly, position of eleventh term from the last term = (24 – 10)^th = 14^th

⇒ 11^th term from the last term = 14^th term from the beginning

Thus, taking n = 14, we need to find a₁₄.

The formula of a_n = a + (n – 1)d

⇒ a₁₄ = 27 + (14 – 1)(-4)

⇒ a₁₄ = 27 – 52 = -25

Hence, the eleventh term from the last term of the AP is -25.