Answer :

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

= 24

∴ -65 is in the 24^{th} position.

To find the eleventh term from the last term,

Position of the 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 a_{14}.

The formula of a_{n} = a + (n – 1)d

⇒ a_{14} = 27 + (14 – 1)(-4)

⇒ a_{14} = 27 – 52 = -25

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

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