Question

If $7$ times the $7^{th}$ term of an Arithmetic Progression is equal to $11$ times its ${11}^{th}$ term, show that its 18th term is zero. Can you find the first term and the common difference? Justify your answer.

Answer 1

We know, $t_n=a+(n-1)d$
Then $t_7=a+(7-1)d$
$\Rightarrow t_7=a+6d$
and $t_{11}=a+(11-1)d$
$\Rightarrow t_{11}=a+10d$
Given that $7t_7=11t_{11}$
Therefore, $7(a+6d)=11(a+10d)$
$\Rightarrow 7a+42d=11a+110d$
$\Rightarrow 11a+110d-7a-42d=0$
$\Rightarrow 4a+68d=0$
$\Rightarrow a+17d=0$
$\Rightarrow a+(18-1)d=0$
$\Rightarrow t_{18}=0$
Impossible to find first term and common difference.

Since these two are unknown but one equation is given.