The 20^{th} term from the end of the arithmet | vedclass

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(D) The given arithmetic progression is $3, 7, 11, \dots, 407$. Here,the first term $a = 3$ and the common difference $d = 7 - 3 = 4$. Let the number

Vedclass · Accepted Answer

$331$

Vedclass · Answer

(D) The given arithmetic progression is $3, 7, 11, \dots, 407$. Here,the first term $a = 3$ and the common difference $d = 7 - 3 = 4$. Let the number of terms be $n$. The $n^{th}$ term is given by $a_n = a + (n - 1)d$. $407 = 3 + (n - 1)4$ $404 = (n - 1)4$ $n - 1 = 101 \implies n = 102$. The $k^{th}$ term from the end is the $(n - k + 1)^{th}$ term from the beginning. Here,$k = 20$ and $n = 102$,so we need the $(102 - 20 + 1)^{th} = 83^{rd}$ term. $a_{83} = a + (83 - 1)d = 3 + 82 	imes 4 = 3 + 328 = 331$.

The $20^{th}$ term from the end of the arithmetic progression $3 + 7 + 11 + \dots + 407$ is ...... .

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