Question

If $7th$ term of $AP$ is $7$ times is equal to $11$ times of $11th$term of $AP$ the that $18th$ term is zero.

Answer 1

Given $7^{th}$ term of A.P is equal to $11$ terms the ${11}^{th}$ term of A.P.

Let $7^{th}$ term of A.P$=a+(n-1)d$

$=a+(7-1)d$

$=a+6d$

d- common difference

a- first term

${11}^{th}$ term of A.P$=a+(11-1)d$

$=a+10d$

Given, $7(a+6d)=11(a+10d)$

$7a+42d=11a+110d$

$\Rightarrow 4a=-68d$

$\Rightarrow a=-17d$ ………..$\left(1\right)$

${18}^{th}$ term of A.P$=a+(18-1)d$

$=a+17d$ [From $\left(1\right)$]

$=-17d+17d$

$=0$.