Question

If the $p^{th}$ term of the series of positive numbers $25,22\frac{3}{5},20\frac{1}{2},18\frac{1}{4}$, .... is numerically the smallest, then the $p^{th}$ is.

• A. $\frac{1}{4}$
• B. $\frac{1}{7}$
• C. $\frac{1}{3}$
• D. $\frac{1}{5}$

Answer 1

The correct option is A $\frac{1}{4}$

$d=22\frac{3}{4}-25$
$=-2\frac{1}{4}=-\frac{9}{4}$
$t_n\ge 0$
$\therefore t_n=25+(n-n)(-\frac{9}{4})$
$25-\frac{9}{4}(n-1)\ge 0$
$\Rightarrow 25\ge \frac{9}{4}(p-1)\ge 0$
$\Rightarrow \frac{100}{9}\ge p-1$
$\Rightarrow \frac{109}{9}\ge p$
$12\frac{1}{9}\ge p$
12th term is positive.
$25+(12-1)\times (-\frac{9}{4})=\frac{1}{4}$