Question

$1\frac{1}{3}+2\frac{1}{4}+3\frac{1}{6}+4\frac{1}{12}+...+19\frac{2}{3}$ is equal to

• A. 220.50
• B. 270.55
• C. 190.95
• D. 210.75

Answer 1

The correct option is A 220.50

$1\frac{1}{3}+2\frac{1}{4}+3\frac{1}{6}+4\frac{1}{12}+...+19\frac{2}{3}$
$=\frac{4}{3}+\frac{9}{4}+\frac{19}{6}+\frac{49}{12}+.....+\frac{59}{3}$
Now, $\frac{9}{4}-\frac{4}{3}=\frac{11}{12}$ and $\frac{19}{6}-\frac{9}{4}=\frac{11}{12}$
Since the difference is same, hence it is an AP
Now, $d=\frac{11}{12},a=\frac{4}{3}$ and using last term l = a+(n-1)d,
$\frac{59}{3}=\frac{4}{3}+(n-1)\times \frac{11}{12}$
$\Rightarrow (n-1)\times \frac{11}{12}=\frac{59}{3}-\frac{4}{3}=\frac{55}{3}$
$\Rightarrow (n-1)=\frac{55}{3}\times \frac{12}{11}$
$\Rightarrow n=20+1$
$\Rightarrow n=21$
Also, $S_n=\frac{n}{2}(a+l)$
$\Rightarrow S_{21}=\frac{21}{2}(\frac{4}{3}+\frac{59}{3})$
$=\frac{21}{2}\times \frac{63}{3}$
=220.50
Hence the correct answer is option (1)