Question

Find the 10th term from the last term of the AP : 8, 10, 12,...., 126.

Answer 1

$AP:8,10,12,...126$

$a=8,d=2,an=126$

Now $an=a+\left(n-1\right)d$

$126=8+\left(n-1\right)2$

$\Rightarrow 118=\left(n-1\right)2$

$\Rightarrow 59=n-1$

$\Rightarrow n=60$

Now ${10}^{th}$ term from last is ${\left(60-10+1\right)}^{th}$ term

$={51}^{th}$ term from beginning

So, $a_{51}=a+\left(51-1\right)d$

$=8+50\times 2$

$=8+100=108$