Question

Find the ${11}^{th}$term, from the last term of the $A.P$ $10,7,4,\dots \dots .-62$

Answer 1

Find the ${11}^{th}$ term from last term of the AP

$10,7,4...,-62$

$a=10$

$d=7-10=-3$

Lets find which term is $-62$

Let $a_n=-62$

$a+(n-1)d=-62$

Substituting the values

$10+(n-1)(-3)=-62$

$10+n(-3)-1(-3)=-62$

$10-3n+3=-62$

$13-3n=-62$

$-3n=-13-62$

$-3n=-75$

$n=\frac{-75}{-3}$

$n=25$

We need to find out ${11}^{th}$ term from last term.

Last term = ${25}^{th}$

second last term $=25-1={24}^{th}$ term

Third last term $=25-2={23}^{rd}$ term

And so on

so ${11}^{th}$ from last term $=(25-10{)}^{th}$ term

$={15}^{th}$ term.

So, we need to find out $a_{15}$

$a_n=a+(n-1)d$

putting values : $a_{15}=10+(15-1)(-3)$

$a_{15}=10+14(-3)$

$=10-42$

$=-32$

$\therefore {11}^{th}$ term from last term of AP is $-32$