Easy MCQ +4 / -1 PYQ · JEE Mains 2024

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

  1. A $-115$ Correct answer
  2. B $-100$
  3. C $-110$
  4. D $-118$

Solution

<p>$$20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots \ldots,-129 \frac{1}{4}$$</p> <p>This is A.P. with common difference</p> <p>$$\begin{aligned} & d_1=-1+\frac{1}{4}=-\frac{3}{4} \\ & -129 \frac{1}{4}, \ldots \ldots \ldots \ldots . . .19 \frac{1}{4}, 20 \end{aligned}$$</p> <p>This is also A.P. $\mathrm{a}=-129 \frac{1}{4}$ and $\mathrm{d}=\frac{3}{4}$</p> <p>Required term $=$</p> <p>$$\begin{aligned} & -129 \frac{1}{4}+(20-1)\left(\frac{3}{4}\right) \\ & =-129-\frac{1}{4}+15-\frac{3}{4}=-115 \end{aligned}$$</p>

About this question

Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression

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