If the 9th term of an A.P is 0 then prove that 29th term is double of the 19th term
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Given a.p.if a=-2 and d=0 then the a.p.of second term will be:
To prove that the 29th term of an arithmetic progression (A.P) is double the 19th term, given that the 9th term is 0, we can use the formula for the nth term of an A.P, which is:
an = a + (n - 1) * d,
where:
an is the nth term,
a is the first term,
n is the position of the term,
d is the common difference.
Let's start by finding the values of a and d using the information given.
Given that the 9th term is 0, we have:
a + (9 - 1)d = 0,
a + 8d = 0. --(1)
Now, to prove that the 29th term is double the 19th term, we need to show that:
a + 28d = 2(a + 18d).
Let's solve for a + 18d on the right side of the equation using equation (1):
a = -8d. --(2)
Substituting equation (2) into the equation we want to prove:
-8d + 28d = 2(-8d + 18d),
20d = 20d.
Since both sides of the equation are equal, this proves that the 29th term is indeed double the 19th term in the given A.P.
This is how we can mathematically prove that the 29th term of an A.P is double the 19th term when the 9th term is 0.