Question

Find the $10$th term from the end of the A.P $810,12,......,126$.

Answer 1

Given sequence is $8,10,12,...,126$

The first term is $a=8$

The common difference is $d=10-8=2$

The last term is $126$

we know that the $n^{th}$ term of the arithmetic progression is given by $a+(n-1)d$

Therefore, $a+(n-1)d=126$

$⟹8+(n-1)2=126$

$⟹2n-2=118$

$⟹2n=120$

$⟹n=\frac{120}{2}=60$

Therefore, there are $60$ terms in the sequence.

Hence, the ${10}^{th}$ term from the end is $60-10+1={51}^{rd}$ term from the beginning.

Therefore, the ${51}^{rd}$ term is $a+(51-1)d=8+50\left(2\right)=108$