Question

plzzzzzzzzzzzz. Help !

Answer 1

Hi !
given :-

a₉ = 0
i.e :-
a +8d = 0        
a = -8d 

To prove :-
a₂₉ = 2(a₁₉)

Proof:-

a₂₉ = a + (29-1)d
      = a + 28d                       
      = -8d + 28d                                            [ ∵ a = -8d ]
      = 20d          ----> (1) 

a₁₉ = a + (19-1)d
     = a + 18d
     = -8d + 18d
    = 10d       -->(2)

From 1 and 2 ,

a₂₉ = 2(a₁₉)

Answer 2

$Given, \\\text{9th term} = 0\\or\ a_9 = 0\\or \ a+8d = 0\\or\ a = -8d.........(1) \\\\Using , (1) \\a_{29} = a + 28\\or \ a_{29} = -8d + 28d\\a_{29} = 20d.........(2)\\\\a_{19} = a+18d\\a_{19} = -8d + 18d\\a_{19} = 10d.........(3)\\\\Dividing \ equation \ (3)by (2):\\\frac{a_{29}}{a_{19}} = \frac{20d}{10d} = 2\\\therefore a_{29} = 2a_{19}\\\\Hence \ Proved!$