Question

Find the ${10}^{th}$ term from end for the A.P. $3,6,9,12,......300$

Answer 1

$3,6,9,\mathrm{12.....300}$

the $AP:a=3d=5$

we need ${10}^{th}$ term from last.

so we treat last term as first term and reverse the series.

$300,297,....12,9,6,3.$

$a=300,d=-3$

${10}^{th}term=a+(n-1)d$

$=300+(10-1)(-3)$

$=300-27$

$=273$

$\therefore {10}^{th}$ term for the end is $273$.