Question

if 9th term of an ap is zero prove that it's 29 term is double the 19th term

Answer 1

Hey

Here is your answer,

Let the first term, common difference and number of terms of an AP are a, d and n respectively.
Given that, 9th term of an AP, T9 = 0 [∵ nth term of an AP, Tn = a + (n-1)d]
⇒ a + (9-1)d = 0
⇒ a + 8d = 0 ⇒ a = -8d ...(i)
Now, its 19th term , T19 = a + (19-1)d
= - 8d + 18d [from Eq.(i)]
= 10d ...(ii)
and its 29th term, T29 = a+(29-1)d
= -8d + 28d [from Eq.(i)]
= 20d = 2 × T19
Hence, its 29th term is twice its 19th term

Hope it helps you!

Answer 2

HERE IS YOUR ANSWER !!!

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Let a and b be the first term and common difference of the given A.P.

9th term of A.P., a9 = 0 [given]

⇒ a + (9 – 1) b = 0

⇒ a + 8b = 0   - - - - -(1)

19th term of A.P. , 

a19= a + (19 – 1) b 
= a + 18b
= a + 8b + 10b 
= 0 + 10b = 10b    (from (1))

∴ 2 a19 =  2 × 10b = 2b    - - - - (2)

29th term of A.P.,  
a29 = a + (29 – 1)b
= a + 28b
= a + 8b + 20b
= 0 + 20b = 20b            (From (1))

∴ a29 =  20b       - - - - - (3)                

SO, 29th term of the given A.P. is double the 19th term of the given A.P.

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hope it helps u !

# Nikky