The fourth ,sixth,ninth terms of AP forms first three terms of GP if first three of AP is 3 find common difference
Let the fourth term of the AP be a, the sixth term be c, and the ninth term be e.
Since the first three terms of the AP are 3, the second term (a1) is 3 - d, the third term (a2) is 3 - 2d, and the fourth term (a3) is 3 - 3d.
We know that a3 = a + 3d, so 3 - 3d = 3 + 3d.
Simplifying this equation, we get -6d = 0, so d = 0.
Therefore, the common difference of the AP is 0.
Let's first find the terms of the arithmetic progression (AP) using the given information:
The first term of the AP is 3.
Let the common difference of the AP be 'd'.
The fourth term of the AP is given by: 3 + 3d
The sixth term of the AP is given by: 3 + 5d
The ninth term of the AP is given by: 3 + 8d
Now, we are told that the first three terms of the AP form a geometric progression (GP). The terms of the GP are:
First term (a) = 3
Common ratio (r) = (3 + 5d) / (3)
The first three terms of the GP are: 3, 3*(3 + 5d) / (3), 3*(3 + 5d) / (3)^2
We can write the equation for the first three terms in terms of the GP as:
3 + 3d = 3
(3 + 5d) / 3 = (3 + 5d) / (3)^2
(3 + 8d) / (3 + 5d) = (3 + 5d) / (3)
Now, let's solve these equations:
From the first equation, we have: 3d = 0
Therefore, the common difference of the AP is d = 0.
However, this would make the second and third terms of the AP the same, which is not possible in an arithmetic progression.
Hence, there is no valid common difference for the given conditions.