JEE MAIN - Mathematics (2024 - 27th January Evening Shift - No. 9)

$$\text { The } 20^{\text {th }} \text { term from the end of the progression } 20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots,-129 \frac{1}{4} \text { is : }$$
$$-115$$
$$-100$$
$$-110$$
$$-118$$

Explanation

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

This is A.P. with common difference

$$\begin{aligned} & d_1=-1+\frac{1}{4}=-\frac{3}{4} \\ & -129 \frac{1}{4}, \ldots \ldots \ldots \ldots . . .19 \frac{1}{4}, 20 \end{aligned}$$

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

Required term $$=$$

$$\begin{aligned} & -129 \frac{1}{4}+(20-1)\left(\frac{3}{4}\right) \\ & =-129-\frac{1}{4}+15-\frac{3}{4}=-115 \end{aligned}$$

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