Question

If 9th term of an

a.P. Is zero, prove that its 29th term is double the 19th temm

Answer 1

Given, 9th term = 0

       a + ( 9 - 1 )d = 0

               a + 8d = 0

                a = - 8d   ---: ( 1 )

Therefore,

     29th term = a + ( 29 - 1 ) d

                       = a + 28d

From ( 1 ), value of a = - 8d      

                       = -8d + 28d

                       = 20 d

19th term = a + ( 19 - 1 )d

                 = a + 18d

From ( 1 ) , value of a = - 8d

                = -8d + 18d

                = 10d

Twice of 19th term = 2 ( 10d )

                                 = 20d

Therefore,

20d =  20d

29 th term = 2( 19th term )

Proved.

$\:$

Answer 2

$\huge \bold {\mathfrak {hey}}$

$\mathbb {\huge {\fcolorbox{red}{pink}{Solution :-}}}$

Given :

9th term = 0

➡️ a + 9 - 1 = d
➡️ a + 8d = 0
➡️ $\bold {a = - 8d - - -(1)}$

Now,
29th term = a + 29 - 1 d
➡️ a + 28d

From equation 1 value pf a = - 8d

➡️ - 8d + 28d
➡️ 20d

19th term = a + (19 - 1) d
➡️ a + 18d

From value of equation 1 a = - 8d

➡️ - 8d + 18d
➡️ 10d

Now,

Twice of 19th term

➡️ 2 ( 10d)
➡️ 20d

Therefore,

20d = 20d

29th term = 2 ( 19th term)

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$\huge \bold {\boxed {Hence, \: Proved}}$

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