If the $ 9^{\text {th }} $ term of an AP is zero, prove that its $ 29^{\text {th }} $ term is twice its $ 19^{\text {th }} $ term.
Given:
The \( 9^{\text {th }} \) term of an AP is zero.
To do:
We have to prove that its \( 29^{\text {th }} \) term is twice its \( 19^{\text {th }} \) term.
Solution:
Let $a$ be the first term and $d$ be the common difference.
This implies,
$a_{9}=a+(9-1)d$
$0=a+8d$
$a=-8d$........(i)
$a_{29}=a+(29-1)d$
$=a+28d$
$=-8d+28d$ [From (i)]
$=20d$.........(ii)
$a_{19}=a+(19-1)d$
$=a+18d$
$=-8d+18d$ [From (i)]
$=10d$........(iii)
Therefore,
$a_{29}=2(10d)$
$=2\times a_{19}$ [From (iii)]
Hence proved.
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