The 20^th term from the end of the progression 20,19(1 / 4),18(1 / 2),17(3 / 4), ... , - 129(1 / 4) is :

20,19 (1 / 4), 18 (1 / 2), 17 (3 / 4), ... ... .,-129 (1 / 4) This is A.P. with common difference \ d_1=-1+(1 / 4)=-(3 / 4) -129 (1 / 4), ... ... ... ... ... 19 (1 / 4), 20 This is also A.P. a=-129 (1 / 4) and d=(3 / 4) Required term =-129 (1 / 4)+(20-1)((3 / 4)) =-129-(1 / 4)+15-(3 / 4) =-115

The 20^{{th}{~}} term from the end of the progression ... | Sequence & Series (Mathematics)

$20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots \ldots .,-129 \frac{1}{4} \\$

This is A.P. with common difference

$\\ \mathrm{d}_1=-1+\frac{1}{4}=-\frac{3}{4}$

$\\-129 \frac{1}{4}, \ldots \ldots \ldots \ldots \ldots 19 \frac{1}{4}, 20 \\\\$

This is also A.P. $\mathrm{a}=-129 \frac{1}{4}$ and $\mathrm{d}=\frac{3}{4}\\$

Required term = $-129 \frac{1}{4}+(20-1)\left(\frac{3}{4}\right)$ $\\=-129-\frac{1}{4}+15-\frac{3}{4} \\$

=-115

Explanation

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