Suppose we have an arithmetic progression a_1 | vedclass

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(A) Given the arithmetic progression $a_1, a_2, \ldots, a_n$ with $a_1 = 1$ and common difference $d = a_2 - a_1 = 5$.The $n$-th term is given by $a_n

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$1011$

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(A) Given the arithmetic progression $a_1, a_2, \ldots, a_n$ with $a_1 = 1$ and common difference $d = a_2 - a_1 = 5$.The $n$-th term is given by $a_n = a_1 + (n - 1)d = 1 + (n - 1)5 = 5n - 4$.We are given $a_k \leq 2021$,so $5k - 4 \leq 2021$,which implies $5k \leq 2025$,or $k \leq 405$.Since $a_{k+1} > 2021$,the sequence is $a_1, a_2, \ldots, a_{405}$.The number of terms is $405$,which is odd. The median of a sequence with $n$ terms (where $n$ is odd) is the term at position $\frac{n+1}{2}$.Here,the median is the term at position $\frac{405+1}{2} = 203$.The $203$-rd term is $a_{203} = a_1 + (203 - 1)d = 1 + 202 	imes 5 = 1 + 1010 = 1011$.

Suppose we have an arithmetic progression $a_1, a_2, \ldots, a_n, \ldots$ with $a_1 = 1$ and $a_2 - a_1 = 5$. The median of the finite sequence $a_1, a_2, \ldots, a_k$,where $a_k \leq 2021$ and $a_{k+1} > 2021$,is

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