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

1.	If the 9th term of an AP is 0, prove that its 29th term is double the 19th term.
Answer» Prove that: 29^th term is double the 19^th term (i.e. a₂₉ = 2a₁₉) Given: a₉= 0 (Where a = a₁ is first term, a₂ is second term, a_n is n^th term and d is common difference of given AP) Formula Used: a_n = a + (n - 1)d So a₉ = 0 → a + (9 - 1)d = 0 a + 8d = 0 a = (–8d) ….equation (i) Now a₂₉ = a + (29 - 1)d and a₁₉ = a + (19 - 1)d a₂₉ = a + 28d and a₁₉ = a + 18d ….equation (ii) By using equation (i) in equation (ii), we have a₂₉= –8d + 28d and a₁₉ = –8d + 18d a₂₉= 20d and a₁₉ = 10d So a₂₉ = 2a₁₉ HENCE PROVED

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

Answer»

Prove that: 29^th term is double the 19^th term (i.e. a₂₉ = 2a₁₉)

Given: a₉= 0

(Where a = a₁ is first term, a₂ is second term, a_n is n^th term and d is common difference of given AP)

Formula Used: a_n = a + (n - 1)d

So a₉ = 0 → a + (9 - 1)d = 0

a + 8d = 0

a = (–8d) ….equation (i)

Now a₂₉ = a + (29 - 1)d and a₁₉ = a + (19 - 1)d

a₂₉ = a + 28d and a₁₉ = a + 18d ….equation (ii)

By using equation (i) in equation (ii), we have

a₂₉= –8d + 28d and a₁₉ = –8d + 18d

a₂₉= 20d and a₁₉ = 10d

So a₂₉ = 2a₁₉

HENCE PROVED