Question

Find the ${10}^{th}$ term from end for the A.P. $3.6.9.\mathrm{12.......300}$

Answer 1

Given, $\mathrm{3.6.9.12.......300}\to A.P$

Here,

first term $\left(a\right)=3$

$n^{th}$ term $\left(Tn\right)=300$

Common difference $\left(d\right)=3$

$w.k.t$, in $A.P.,Tn=a+(n-1)d$

So,

$300=3+(n-1)\times 3$

$297=(n-1)3$

$\frac{297}{3}+1=n$

$\therefore n=100$

So, we have total no. of terms $\left(n\right)=100$ we have to find ${10}^{th}$ term from 'end' which ${91}^{st}$ term from 'beginning'

So,

${91}^{st}$ term from beginning

$T_{91}=a+(91-1)d$

$T_{91}=3+\left(90\right)\times 3$

$T_{91}=3+270$

$T_{91}=273$