Question

If the 9th term of an ap is zero, prove that its 29th term is twice its 19th term

Answer 1

$\huge Hello$$\huge friends,$


Let a and d respectively be the first term and common difference of the AP.

Given a9 = 0

So, a + (9-1)d = 0

a+8d=0

a= -8d

 

Now, 29th term = a+28d

=-8d+28d

= 20d = 2 x 10d

= 2(-8d + 18d)

=2(a+18d)

= 2 x 19th term

 

Thus, the 29th term of the AP is twice the 19th term.





$\huge BeBrainly$

$\huge @techybhai$

Answer 2

Given:

$a_{9} = 0$

We are asked to prove that:

$a_{29} = 2(a_{19})$

Using the formula:

$a_{n} = a + (n - 1)d, \: \: Where \: \: a \: \: is \\ \: \: the \: \: 1st \: \: term, \: \: n \: \: is \: \: the \: \: {n}^{th} \: \: \\ term \: \: and \: \: d \: \: is \: \: the \: \: common \: \: \\ difference.$

Substituting n = 9 in this formula:

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

$a_{29} = a + (29 - 1)d \\ a_{29} = a + 28d = - 8d + 28d \\ a_{29} = 20d$

$a_{19} = a + (19 - 1)d \\ a_{19} = a + 18d = - 8d + 18d \\ a_{19} = 10d$

20d = 2(10d).

20d = 20d.

Therefore, it is proved that the 29th term is twice the 19th term when the 9th term is 0.