Question

If the $9^{th}$ term of as AP is zero, then prove that its ${29}^{th}$ term is twice its ${19}^{th}$ term.

Answer 1

Given:

$a_9=2\left(a_{19}\right)$

We are asked to prove that:

$a_{29}=2\left(a_{19}\right)$

Using the formula:

$a_9=a+(n-1)d$, where a is the 1st term, n is the nth 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.