Question

If the 9th term of an AP is 0,prove that its 29th term is double to 19th term.

Answer 1

here is the answer...

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Answer 2

Given that 9th term of an AP is 0.

Let a be the first term and d be the common difference.

Now, T9 = 0

We know that sum of n terms of an Ap sn = a + (n - 1) * d

                                                                  s9 = a + (9 - 1) * d = 0

                                                                        a + 8d = 0


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

s19 = a + (19 - 1) * d

       = a + 18d

       = (a + 8d) + 10d

       = 0 + 10d

       = 10d.


s29 = a + (29 - 1) * d

       = a + 28d

      = (a + 8d) + 20d

      = 0 + 20d

     = 20d

     = 2 * (10d)

     = 2 * s9.


Hope this helps!