Question

if 9th term of an AP is 0 prove it's 29th term is double the 19th term

Answer 1

t9=0. a+8d=0. a=-8d
t29= a+(n-1)d
=a+28d
=-8d +28d=20d
t19= a + 18d
=a + 28d -8d
= a+ 28d+a
=2a+28d
=2(t29)

Answer 2

Answer:

29th term of given AP is double the 19th term.

Step-by-step explanation:

The nth term of an AP is

$a_n=a+(n-1)d$

It is given that 9th term of an AP is 0.

$a_9=a+(9-1)d$

$0=a+8d$

$a=-8d$

We have to prove that 29th term is double the 19th term.

The 29th term of AP is

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

The 19th term of AP is

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

Since 29th term of AP is 20d and 19th term of AP is 10d. So

$a_{29}=2\times a_{19}$

Hence proved that 29th term is double the 19th term.