Question

The sets $S_1,S_2,S_3.....$ are given by $S_1=\left\{\frac{2}{1}\right\},S_2=\left\{\frac{3}{2},\frac{5}{2}\right\},S_3=\left\{\frac{4}{3},\frac{7}{3},\frac{10}{3}\right\},S_4=\left\{\frac{5}{4},\frac{9}{4},\frac{13}{4},\frac{17}{4}\right\}........$ Then the sum of the numbers in set $S_{25}$ is

• A. $320$
• B. $322$
• C. $324$
• D. $326$

Answer 1

The correct option is D

$326$

Explanation for correct option:

$$\begin{gathered} S_1=\left\{2,1\right\} \\ {S}_2=\left\{\frac{3}{2},\frac{5}{2}\right\} \\ {S}_3=\left\{\frac{4}{3},\frac{7}{3},\frac{10}{3}\right\} \\ {S}_4=\left\{\frac{5}{4},\frac{9}{4},\frac{13}{4},\frac{17}{4}\right\} \end{gathered}$$

The sequence in the sets is $S_n=\left\{1+\frac{1}{n},2+\frac{1}{n},3+\frac{1}{n},......n\mathrm{terms}\right\}$

Which is an arithmetic progression with the first term being $1+\frac{1}{n}$ and common difference $d=1$

So,$S_{25}=\left\{1+\frac{1}{25},2+\frac{1}{25},........25\mathrm{terms}\right\}$

Sum of all elements in $S_{25}=$sum of the arithmetic progression

Sum of arithmetic progression$=\frac{n}{2}\left(2a+(n-1)d\right)$

For $S_{25}$, $n=25$

Sum of all elements in $S_{25}=\frac{25}{2}\left[2\left(1+\frac{1}{25}\right)+(25-1)1\right]$

$$\begin{gathered} \Rightarrow S_{25}=\frac{25}{2}\left(2\times \frac{26}{25}+24\right) \\ \Rightarrow S_{25}=\frac{25}{2}\left(\frac{52}{25}+24\right) \\ \Rightarrow S_{25}=\frac{25}{2}\left(\frac{652}{25}\right) \\ \Rightarrow S_{25}=326 \end{gathered}$$

Hence, option D is correct.