NIMCET 2009 — Reasoning PYQ

1. Number of Elements in Each Set:

• $A_1$ has $1$ element.

• $A_2$ has $3$ elements.

• $A_3$ has $5$ elements.

• The number of elements in $A_n$ follows the sequence $1,3,5,7,\dots$, which is $(2n-1)$.

2. Total Elements Before $A_{20}$:

The total number of terms used in sets $A_1$ through $A_{n-1}$ is the sum of the first $(n-1)$ odd numbers:

For $A_{20}$, the total terms before it are $(20-1{)}^2=19^2=361$ terms.

3. Finding the Terms of $A_{20}$:

The elements are consecutive odd numbers starting from $3$. The $k^{th}$ odd number in this overall sequence is given by $T_k=2k+1$.

• The terms of $A_{20}$ start after the $361^{st}$ term.

• The number of elements in $A_{20}$ is $2(20)-1=39$.

4. Calculating the Average:

For any set of numbers in an Arithmetic Progression (like these odd numbers), the average is simply the middle term.

The middle term of $A_{20}$ is the term that falls exactly in the center of its $39$ elements.

• Position of the middle term in $A_{20}=\frac{39+1}{2}=20^{th}$ element of $A_{20}$.

• Overall position of this term in the entire sequence $=\text{Terms before }{A}_{20}+20$.

• Overall position $=361+20=381^{st}$ term.

5. Final Calculation:

Using the formula $T_k=2k+1$:

Correct Option: (b) 763