Final answer:
To prove that the 19th term of an AP is half of its 29th term given that the 9th term is zero, we use the formula for the nth term of an AP. We rearrange the equation and solve for the first term, which we find is equal to -16 times the common difference. Therefore, a19 = 2d and a29 = 2d, showing that a19 = (1/2) * a29.
Step-by-step explanation:
To prove that the 19th term of an AP is half of its 29th term given that the 9th term is zero, let's use the formula for the nth term of an AP:
an = a1 + (n - 1)d
Where an is the nth term, a1 is the first term, and d is the common difference. We're given a9 = 0, so plugging in the values:
a9 = a1 + (9 - 1)d = 0
a1 + 8d = 0
Similarly, we can find a19 and a29:
a19 = a1 + (19 - 1)d
a29 = a1 + (29 - 1)d
We need to prove that a19 = (1/2) * a29. To do this, let's subtract a1 + 8d from both sides of the equation for a29:
a29 - (a1 + 8d) = 0
By rearranging the equation, we get:
a1 + 18d = 2(a1 + 8d)
Expanding the right side:
a1 + 18d = 2a1 + 16d
Now, by rearranging the equation:
2a1 - a1 = 2d - 18d
a1 = -16d
This equation implies that a1 is equal to -16 times the common difference. Therefore, a19 = (-16d) + (19 - 1)d = 2d and a29 = (-16d) + (29 - 1)d = 2d.
Thus, we have shown that a19 = (1/2) * a29.