If the 9th term of an A.P is 0 then prove that 29th term is double of the 19th term
To prove that the 29th term of the arithmetic progression (A.P) is double the 19th term when the 9th term is 0, we can use the formula for the nth term of an A.P:
nth term (Tn) = a + (n-1)d
where 'a' is the first term and 'd' is the common difference.
Given that the 9th term (T9) is 0, we have:
T9 = a + (9-1)d = 0
Simplifying this equation, we get:
a + 8d = 0 ---(1)
Now, we need to prove that the 29th term (T29) is double the 19th term (T19).
So, we have:
T29 = a + (29-1)d = a + 28d ---(2)
T19 = a + (19-1)d = a + 18d ---(3)
To prove that T29 is double T19, we can show that:
T29 = 2T19
Replacing the values from equations (2) and (3), we have:
a + 28d = 2(a + 18d)
Simplifying this equation, we get:
a + 28d = 2a + 36d
Rearranging the equation, we have:
a - 2a = 36d - 28d
Simplifying further, we get:
-a = 8d
Since we already have the equation (1), which is a + 8d = 0, we can conclude that:
-a = 8d
Therefore, T29 = 2T19.
Hence, the 29th term is double the 19th term in the given arithmetic progression.