If 9th term of an arithmetic progression is 0, prove that its 29th term is double the 19th term.
If the 9th term of an AP is 0 , then
a+8d = 0
a = -8d
29th term = a + 28d
= -8d + 28d = 20d
19th term = a +18d
= -8d + 18d = 10d
since 20d = 2(10d)
the 29th term is twice the 19th term
To prove that the 29th term of an arithmetic progression (AP) is double the 19th term, given that the 9th term is 0, we will use the formula for the nth term of an AP.
The formula for the nth term of an arithmetic progression is given by:
an = a + (n - 1)d
Where:
an is the nth term of the AP,
a is the first term of the AP,
n is the position of the term in the AP, and
d is the common difference between consecutive terms.
Given that the 9th term in the AP is 0, we can rewrite the formula as:
a9 = a + (9 - 1)d = 0
Simplifying this equation gives us:
a + 8d = 0 .......(Equation 1)
Now, we need to find the 19th and 29th terms of the AP and prove that the 29th term is double the 19th term.
For the 19th term:
a19 = a + (19 - 1)d
= a + 18d
For the 29th term:
a29 = a + (29 - 1)d
= a + 28d
To prove that the 29th term is double the 19th term, we need to show that:
a29 = 2(a19)
Substituting the values of a19 and a29:
a + 28d = 2(a + 18d)
Expanding and simplifying this equation:
a + 28d = 2a + 36d
Rearranging the terms:
a - 2a = 36d - 28d
Simplifying further:
-a = 8d .......(Equation 2)
Now, we have Equation 1 (a + 8d = 0) and Equation 2 (-a = 8d).
Adding these two equations:
(a + 8d) + (-a) = 0 + 8d
a - a + 8d = 8d
Simplifying:
8d = 8d
This proves that the 29th term (a29) is double the 19th term (a19) in the given arithmetic progression.
Hence, we have proved that if the 9th term of an arithmetic progression is 0, then the 29th term is double the 19th term.