Question

✨ If the 9 th term of An AP is 0 then prove that its 29 th term is double of its 19 th term ✨

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Answer 1

Let , Common difference = d

First term = a

Given :

$\to$ 9th Term = 0

$\to$ 9th Term => a + 8d = 0

Therefore , a = - 8d

According to the question

We have to prove 29th term is double of its 19th term ( i.e 19th term × 2 = 29th term )

So ,

$\to$ 29th term = a + 28d

$\to$ 29th Term = - 8d + 28d

$\to$ 29th Term = 20d

___________

$\to$ 19th term = a + 18d

$\to$ 19th term = - 8d + 18d

$\to$ 19th term = 10d

Hence Proved

Answer 2

SOLUTION:-

Given:

⚫9th term of an A.P = 0.

Let the first term of an A.P.=a

& common difference= d.

To prove:

We have to prove that,

${}^{t}29 = {}^{2t} 19$

We know that nth term of an A.P.,

${}^{t}n = a + (n - 1)d \\ \\ = > {}^{t} 9 = a + (9 - 1)d \\ \\ = > {}^{t}9 = a + 8d$

But, t9= 0

=) a+8d= 0

=) a= -8d............(1)

Now,

${}^{ t }29 = a + (29 - 1)d \\ \\ = > a + 28d$

=) -8d + 28d [from (1)]

=) 20d

Therefore,

t29= 20d.............(2)

Now,

${}^{t} 19 = a + (19 - 1)d \\ \\ = > a + 18d$

=) -8d + 18d [from (1)]

=) 10d

Therefore,

t19= 10d.............(3)

Equating (2) & (3), we get:

=)20d = 10d

=)20d = 10d(2)

Therefore,

${}^{t}29= 2( {}^{t} 19)$

Hence, proved

Hope it helps ☺️