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In an arithmetic progression, if the $k^{th}$ term is $5k+1$, then find the sum of first $100$ terms.
Given: In an arithmetic progression, $k^{th}$ term is $5k+1$.
To do: To find the sum of first $100$ terms.
Solution:
Let $a$ be the first term of an AP and $d$ is the common difference.
$\therefore a_k=a+(n-1)d$
Since, $a_k=5k+1$
$a+( k-1)d=5( k-1)+6$
$\Rightarrow a+( k-1)d=6+( k-1)5$
Equating both sides, we get
$a=6$ and $d=5$
$\therefore$ Sum of $100$ terms, $S_{100}=\frac{n}{2}[2a+( n-1)d]$
$=\frac{100}{2}[2\times6+99\times5]$
$=50[12+495]=50( 507)=25, 350$
Thus, the sum of $100$ terms is $25, 350$.
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