determine the ap whose third term is 16 and 7th term exceeds the 5th term by 12
Determine the arithmetic progression (AP) given that the third term is 16 and the 7th term exceeds the 5th term by 12
Answer:
Let’s denote the first term of the arithmetic progression as a, and the common difference as d.
We know that the third term of an arithmetic progression is given by a + 2d and the seventh term is given by a + 6d.
Given that the third term is 16, we have:
a + 2d = 16 …(1)
Given that the 7th term exceeds the 5th term by 12, we have:
(a + 6d) - (a + 4d) = 12
2d = 12
d = 6
Substitute d = 6 into equation (1):
a + 2(6) = 16
a + 12 = 16
a = 4
Therefore, the arithmetic progression is:
4, 10, 16, 22, 28, 34, …