Question

If 9th term of an A.P. is zero, prove that its 29th term is double the 19th term.

Answer 1

Given:
$a_9=0$
$\therefore a+8d=0a=-8d\dots \dots \left(i\right)a_{19}=a+(19-1)d=a+18d$
$=-8d+18d$ $[\because a=-8dfrom(i\left)\right]$
$=10d\dots \dots \left(ii\right)$
$a_{29}=a+(29-1)d$
$=-8d+28d$ $[\because a=-8dfrom(i\left)\right]$
$=20d\dots \dots \left(iii\right)$
From (ii) and (iii), we can conclude
$a_{29}=2a_{19}$
Hence proved.