x-2, 5 and x-2 are the 2nd, 3rd and 4th term of an ap respectively. find x
Are the second and fourth terms supposed to be the same?
YES
Good question. If so, then the common difference must be zero, so
x-2 = 5
x = 7
and the terms are 5,5,5
However, if the last terms is supposed to be x+2, then the terms are clearly
3,5,7
Pls I need solution to this question
To find the value of x, we can use the formula for the nth term of an arithmetic progression (AP):
nth term (Tn) = a + (n-1)d
Here, a represents the first term of the AP, d represents the common difference, and n represents the position of the required term.
Given that the second term is x-2, the third term is 5, and the fourth term is x-2, we can substitute these values into the formula. Let's find the common difference first:
Second term: T2 = a + (2-1)d = x-2
Third term: T3 = a + (3-1)d = 5
Fourth term: T4 = a + (4-1)d = x-2
We can now set up two equations using the given information:
Equation 1: T3 = 5
5 = a + 2d
Equation 2: T4 = x-2
x-2 = a + 3d
Now we have a system of equations with two variables (a and d). We can solve these equations simultaneously to find the values of a and d.
Subtracting Equation 1 from Equation 2, we get:
(x-2) - 5 = (a + 3d) - (a + 2d)
x - 7 = d
Now we substitute this value of d into Equation 1:
5 = a + 2(x - 7)
5 = a + 2x - 14
19 = a + 2x
Now we have two equations:
Equation 3: 19 = a + 2x
Equation 4: x - 7 = d
We can simplify Equation 3 to solve for a:
a = 19 - 2x
Substituting this value of a into Equation 4, we get:
x - 7 = d
Since we already know that d = x - 7, we can substitute this into the equation:
x - 7 = x - 7
This equation is true for any value of x. Therefore, x can take any value.
In conclusion, the value of x can be any real number.