Question

Find the ${17}^{th}$ term from the end of the AP $1,6,11,\mathrm{16...201},216$.

Answer 1

$a=1d=5$

Number of terms when $216$ is the last term.

$1+5(n-1)=216$

$(n-1)=\frac{215}{5}=43,n=44$

${17}^{th}$ term from last $=44-17={27}^{th}$ term.

${27}^{th}$ term $=1+(27-1)\times 5=1+26\times 5=131$