Question

If $9^{th}$ term of an $A.P$, is zero, then prove that its ${29}^{th}$ term is double than the ${19}^{th}$ term.

Answer 1

$9th$ term, $a+8n=0a=-8n$

$29th$ term, $a+28n=(-8+28)n=20n$

$19th$ term $a+18n=(-8+18)n=10n$

Hence $29th$term is twice of $19th$ term.