Let Sn(n≥1) be a sequence of sets defined by S1={0},S2={32,52},S3={83,113,143},S4={154,194,234,274},...., then

The 1st elements of s1,s2,s3,s4,...., are 12−11,22−12,32−13,....si must be a set starting at t2−1i s20={39920,399+2020,399+4020,...20 term}⇒ third term is 43920 s20=202[2×39920+19×1]=589.

Let Sn(n≥1) be a sequence of sets defined by S1={0},S2={32,52},S3={83,113,143},S4={154,194,234,274},...., then

The 1st elements of s1,s2,s3,s4,...., are 12−11,22−12,32−13,....si must be a set starting at t2−1i s20={39920,399+2020,399+4020,...20 term}⇒ third term is 43920 s20=202[2×39920+19×1]=589.

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Let $S_{n} (n \geq 1)$ be a sequence of sets defined by $S_{1} = {0}, S_{2} = {\frac{3}{2}, \frac{5}{2}}, S_{3} = {\frac{8}{3}, \frac{11}{3}, \frac{14}{3}}, S_{4} = {\frac{15}{4}, \frac{19}{4}, \frac{23}{4}, \frac{27}{4}}, ....,$
then

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third element in $S_{20}$ is $\frac{439}{20}$

sum of elements in $S_{20}$ is 609

third element in $S_{20}$ is $\frac{431}{20}$

sum of elements in $S_{20}$ is 589

answer is A, C.

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Detailed Solution

The 1st elements of $s_{1}, s_{2}, s_{3}, s_{4}, ....,$ are $\frac{1^{2} - 1}{1}, \frac{2^{2} - 1}{2}, \frac{3^{2} - 1}{3}, ....$
$s_{i}$ must be a set starting at $\frac{t^{2} - 1}{i}$
$s_{20} = {\frac{399}{20}, \frac{399 + 20}{20}, \frac{399 + 40}{20}, ...20 t e r m}$
$\Rightarrow$ third term is $\frac{439}{20}$
$s_{20} = \frac{20}{2} [\frac{2 \times 399}{20} + 19 \times 1] = 589.$