The 4^{text{th}} term from the end of the AP: | vedclass

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(C) The $n^{	ext{th}}$ term of an $AP$ from the end is given by the formula:$a_n = l - (n - 1)d$Where $l$ is the last term,$n$ is the position from t

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(C) The $n^{	ext{th}}$ term of an $AP$ from the end is given by the formula:$a_n = l - (n - 1)d$Where $l$ is the last term,$n$ is the position from the end,and $d$ is the common difference.Given the $AP: -11, -8, -5, \dots, 49$,the last term $l = 49$.The common difference $d = -8 - (-11) = -8 + 11 = 3$.We need to find the $4^{	ext{th}}$ term from the end,so $n = 4$.Substituting the values into the formula:$a_4 = 49 - (4 - 1) 	imes 3$$a_4 = 49 - (3 	imes 3)$$a_4 = 49 - 9 = 40$.Thus,the $4^{	ext{th}}$ term from the end is $40$.

The $4^{\text{th}}$ term from the end of the $AP: -11, -8, -5, \dots, 49$ is

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