Question

For an A.P. given below find $t_{20}$ and $S_{10}$.
$\frac{1}{6},\frac{1}{4},\frac{1}{3},...$

• A. $\frac{7}{4},\frac{65}{12}$
• B. $\frac{5}{4},\frac{63}{12}$
• C. $\frac{5}{4},\frac{65}{12}$
• D. $\frac{7}{4},\frac{63}{12}$

Answer 1

The correct option is A $\frac{7}{4},\frac{65}{12}$

Here first term $a=\frac{1}{6}$ and common difference $d=\frac{1}{4}-\frac{1}{6}=\frac{1}{12}.$
Now, $t_{20}=a+19d\Rightarrow t_{20}=\frac{1}{6}+19\times \frac{1}{12}=\frac{21}{12}=\frac{7}{4}$
Also $S_{10}=\frac{10}{2}[2a+(10-1\left)d\right]$
$\Rightarrow S_{10}=5[2\times \frac{1}{6}+9\times \frac{1}{12}]=5[\frac{1}{3}+\frac{3}{4}]=\frac{65}{12}$