Question

If the 9th term of an AP is 0, Prove that its 29th term is twice its 19th term.


No Spamming please.

Answer 1

$\huge{ \bold{Solution: }}$
Arithmetic Sequence: A Sequence in which each term differ from it's preceding term by a constant Number is called AP.



$\huge{ \bold{Given: }}$

T9=a+8d

•°• 0 =a+8d


$\huge{ \bold{To \: Prove - }}$
Prove that its 29th term is twice its 19th term.

Answer 2

Answer:

Step-by-step explanation:

We know that:nth term of an AP=

a+(n-1)d,where a is the first term and d is the common difference.

Here,a9=0

=>a+8d=0=>;a= -8d

Now,29th term a29=a+28d= -8d+28d=20d1

9th term a19=a+18d= -8d+18d=10d

So,29th term is twice of 19th term.