If 9th term of an A.P. is zero, prove its 29th term is double the 19

1.	If 9th term of an A.P. is zero, prove its 29th term is double the 19th term.
Answer» Given, a₉ = 0 We know that, n^th term a_n = a + (n – 1)d So, a + (9 – 1)d = 0 ⇒ a + 8d = 0 ……(i) Now, 29^th term is given by a₂₉ = a + (29 – 1)d ⇒ a₂₉ = a + 28d And, a₂₉ = (a + 8d) + 20d [using (i)] ⇒ a₂₉ = 20d ….. (ii) Similarly, 19^th term is given by a₁₉ = a + (19 – 1)d ⇒ a₁₉ = a + 18d And, a₁₉ = (a + 8d) + 10d [using (i)] ⇒ a₁₉ = 10d …..(iii) On comparing (ii) and (iii), it’s clearly seen that a₂₉ = 2(a₁₉) Therefore, 29^th term is double the 19^th term.

If 9th term of an A.P. is zero, prove its 29th term is double the 19th term.

Answer»

Given,

a₉ = 0

We know that, n^th term a_n = a + (n – 1)d

So, a + (9 – 1)d = 0 ⇒ a + 8d = 0 ……(i)

Now,

29^th term is given by a₂₉ = a + (29 – 1)d

⇒ a₂₉ = a + 28d

And, a₂₉ = (a + 8d) + 20d [using (i)]

⇒ a₂₉ = 20d ….. (ii)

Similarly, 19^th term is given by a₁₉ = a + (19 – 1)d

⇒ a₁₉ = a + 18d

And, a₁₉ = (a + 8d) + 10d [using (i)]

⇒ a₁₉ = 10d …..(iii)

On comparing (ii) and (iii), it’s clearly seen that

a₂₉ = 2(a₁₉)

Therefore, 29^th term is double the 19^th term.