Question

The smallest positive term of the sequence $25,22\frac{3}{4},20\frac{1}{2},18\frac{1}{4},\cdots$ is

• A. $7$
• B. $\frac{19}{4}$
• C. $\frac{5}{2}$
• D. $\frac{1}{4}$

Answer 1

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

$25,22\frac{3}{4},20\frac{1}{2},18\frac{1}{4},\cdots$
$\Rightarrow \frac{100}{4},\frac{91}{4},\frac{82}{4},\frac{73}{4},\cdots$

The sequence is an A.P., where $a=25,d=-\frac{9}{4}$
Now, $T_n=a+(n-1)\times d$
$\Rightarrow T_n=25+(n-1)\times (-\frac{9}{4})\Rightarrow T_n=25+\frac{9}{4}-\frac{9n}{4}\Rightarrow T_n=\frac{109}{4}-\frac{9n}{4}$

When $T_n$ is negative,
$\frac{109}{4}-\frac{9n}{4}<0\Rightarrow n>\frac{109}{9}$
As $\frac{109}{9}>12$, so $T_{13}$ will be the first negative term.
Therefore, $T_{12}$ will be the smallest positive term.
$T_{12}=25+(12-1)\times (-\frac{9}{4})\Rightarrow T_{12}=\frac{1}{4}$

Hence, the smallest positive term is $\frac{1}{4}$.