Find the $17^{th}$ term from the end of the AP: $1,\ 6,\ 11,\ 16\ …201,\ 216$.
Given: AP: $1,\ 6,\ 11,\ 16\ …201,\ 216$.
To do: To find the $17^{th}$ term from the end of the AP.
Solution:
$a=1,\ d=5$
Number of terms if last term is $216$.
$1+( n−1)\times5=216$
$\Rightarrow 5( n−1)=216-1=215$
$\Rightarrow n-1=\frac{215}{5}$
$\Rightarrow n-1=43$
$\Rightarrow n=44$
$17^{th}$ term from end $=44−17=27^{th}$ term from first,
$27^{th}$ term $=1+( 27−1)\times5$
$=1+26\times5$
$=131$
The $17^{th}$ term from the end of the AP is $131$.
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