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

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(C) The given progression is an $A.P.$ with first term $a = 20$ and common difference $d = 19 \frac{1}{4} - 20 = -\frac{3}{4}$.To find the $n^{	ext{t

Vedclass · Accepted Answer

$-115$

Vedclass · Answer

(C) The given progression is an $A.P.$ with first term $a = 20$ and common difference $d = 19 \frac{1}{4} - 20 = -\frac{3}{4}$.To find the $n^{	ext{th}}$ term from the end,we can reverse the $A.P.$The new $A.P.$ starts from $-129 \frac{1}{4}$ with a common difference $d' = -d = \frac{3}{4}$.The $n^{	ext{th}}$ term from the end is given by $a_n = a_{last} + (n-1)d'$.Here,$a_{last} = -129 \frac{1}{4} = -\frac{517}{4}$,$n = 20$,and $d' = \frac{3}{4}$.$a_{20} = -\frac{517}{4} + (20-1) 	imes \frac{3}{4}$$a_{20} = -\frac{517}{4} + 19 	imes \frac{3}{4}$$a_{20} = \frac{-517 + 57}{4} = \frac{-460}{4} = -115$.

The $20^{\text{th}}$ term from the end of the progression $20, 19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots, -129 \frac{1}{4}$ is:

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