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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Updated on: 10-Oct-2022

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