In A P the third term is 4 the first term The 6th term is 17 find series
I assume you meant to say: the third term is four times the first term
---> a+2d = 4a
2d = 3a
d = 3a/2
also, a+5d = 17
a + 5(3a/2) = 17
2a + 15a = 34
17a = 34
a = 2 , then d = 3
series is 2+5+8+11+ ...
To find the series in an arithmetic progression (AP), we need to determine the common difference and the first term.
Given information:
- Third term (A3) = 4
- First term (A1)
- Sixth term (A6) = 17
Let's use the formula for the nth term in an AP: An = A1 + (n-1)d
We have two equations:
- A3 = A1 + 2d (since the third term is A3 and the common difference is d)
- A6 = A1 + 5d (since the sixth term is A6)
Substituting the values into the equations, we have:
4 = A1 + 2d ----(1)
17 = A1 + 5d ----(2)
To solve this system of equations, we can use the method of substitution or elimination.
Method of substitution:
From equation (1), we can express A1 in terms of d:
A1 = 4 - 2d
Substituting A1 in equation (2):
17 = (4 - 2d) + 5d
Simplifying the equation:
17 = 4 + 3d
13 = 3d
d = 13/3
Now that we have the value of the common difference (d), we can substitute it back into equation (1) to find A1:
4 = A1 + 2(13/3)
Simplifying the equation:
4 = A1 + (26/3)
4 - 26/3 = A1
(12/3) - (26/3) = A1
-14/3 = A1
Therefore, the first term (A1) of the arithmetic progression is -14/3, and the common difference (d) is 13/3.
The series can be represented as:
-14/3, -8/3, -2/3, 4/3, 10/3, 16/3, ...
Note that this series continues indefinitely, with each term being obtained by adding the common difference (d) to the previous term.