**Category : **JEE Main & Advanced

(1) Let \['a'\] be the first term and \['d'\] be the common difference of an A.P. Then its \[{{n}^{th}}\] term is \[a+(n-1)d\]*i.e.*, \[{{T}_{n}}=a+(n-1)d\].

(2) \[{{p}^{th}}\] **term of an A.P. from the end : **Let \['a'\] be the first term and \['d'\] be the common difference of an A.P. having \[n\] terms. Then \[{{p}^{th}}\] term from the end is \[{{(n-p+1)}^{th}}\] term from the beginning

* *

*i.e.*, \[{{p}^{th}}\text{ term from the end }=\text{ }{{T}_{(n-p+1)}}=a+(n-p)d\].

- If last term of an A.P. is
*l*then \[{{p}^{th}}\]term from end\[=l-(p-1)d\]

*play_arrow*Definition*play_arrow*General Term of an A.P.*play_arrow*Selection of Terms in an A.P.*play_arrow*Sum of n terms of an A.P.*play_arrow*Arithmetic Mean*play_arrow*Properties of A.P.

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