The 8th term of an AP is 38 and the 12rh term is 58.find the 30th term
a+7d=38
a+11d=58
subtract
4d = 20
d=5
backsub
a+35=38
a = 3
now use your formulas I use to find term30
Tn=a+(n-1)d,3+(30-1)*5=148
To find the 30th term of an arithmetic progression (AP), we need to determine the common difference first.
Given:
8th term (a8) = 38
12th term (a12) = 58
Step 1: Find the common difference (d) using the formula a(n) = a(1) + (n - 1)d.
We can write two equations using the given terms.
Using the 8th term:
a(8) = a(1) + (8 - 1)d
38 = a(1) + 7d ----(Equation 1)
Using the 12th term:
a(12) = a(1) + (12 - 1)d
58 = a(1) + 11d ----(Equation 2)
Step 2: Solve the system of equations created in Step 1 to find the values of a(1) and d.
Subtracting Equation 1 from Equation 2, we get:
58 - 38 = a(1) + 11d - a(1) - 7d
20 = 4d
d = 20 / 4
d = 5
Substituting the value of d into Equation 1, we can solve for a(1):
38 = a(1) + 7(5)
38 = a(1) + 35
a(1) = 38 - 35
a(1) = 3
Step 3: Find the 30th term (a30) using the formula a(n) = a(1) + (n - 1)d.
a(30) = a(1) + (30 - 1)d
a(30) = 3 + 29(5)
a(30) = 3 + 145
a(30) = 148
Therefore, the 30th term of the arithmetic progression is 148.
To find the 30th term of the arithmetic progression (AP), we need to first find the common difference (d) between consecutive terms.
The formula to find the nth term in an AP is as follows:
An = a + (n - 1)d
where An is the nth term, a is the first term, n is the term number, and d is the common difference.
Given that the 8th term is 38, we can substitute the values into the formula to get:
38 = a + 7d ----(1)
Similarly, the 12th term is given as 58, so we have:
58 = a + 11d ----(2)
We now have a system of two equations with two variables (a and d).
Subtracting equation (1) from equation (2), we get:
58 - 38 = (a + 11d) - (a + 7d)
20 = 4d
d = 20/4
d = 5
Now that we have the common difference (d = 5), we can use the formula to find the 30th term:
A30 = a + (30 - 1)d
A30 = a + 29(5)
A30 = a + 145
To find the value of 'a', we can substitute the common difference (d = 5) into equation (1):
38 = a + 7(5)
38 = a + 35
a = 38 - 35
a = 3
Now we can substitute the values of a and d into A30 to find the 30th term:
A30 = 3 + 145
A30 = 148
Therefore, the 30th term of the given arithmetic progression is 148.