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