Question:
The $9^{\text {th }}$ term of an AP is 0 . Prove that its $29^{\text {th }}$ term is double the $19^{\text {th }}$ term.
Solution:
Given : $9^{\text {th }}$ term is 0
To prove: $29^{\text {th }}$ term is double the $19^{\text {th }}$ term
$a+8 d=0$
$a=-8 d$
$29^{\text {th }}$ term is
$a+28 d$
$\Rightarrow 20 d$
$19^{\text {th }}$ term is
$a+18 d$
$\Rightarrow 10 d$
Hence proved $29^{\text {th }}$ term is double the $19^{\text {th }}$ term
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