The 7th term of the A.P is 34 and 15th term is 74. Find it's 40th term.
In an A.p.,t7=34 and t15=74 then t40 =
To find the 40th term of an arithmetic progression (AP), we need to determine the common difference first.
Given information:
The 7th term (a[7]) is 34.
The 15th term (a[15]) is 74.
Since the formula for the nth term of an AP is given by a[n] = a[1] + (n-1)d, we can use this information to set up two equations:
a[7] = a[1] + 6d
34 = a[1] + 6d .....(equation 1)
a[15] = a[1] + 14d
74 = a[1] + 14d ......(equation 2)
We have two equations and two unknowns (a[1] and d), so we can solve this system of equations simultaneously to find the values of a[1] and d.
Subtracting equation 1 from equation 2, we get:
74 - 34 = (a[1] + 14d) - (a[1] + 6d)
40 = 14d - 6d
40 = 8d
Dividing both sides by 8, we find:
d = 5
Now that we have found the common difference (d = 5), we can substitute this value into equation 1 to find a[1].
Using equation 1:
34 = a[1] + 6d
34 = a[1] + 6(5)
34 = a[1] + 30
a[1] = 34 - 30
a[1] = 4
Now that we know the first term (a[1] = 4) and the common difference (d = 5), we can find the 40th term (a[40]) using the formula:
a[n] = a[1] + (n-1)d
Substituting the values:
a[40] = 4 + (40 - 1)5
a[40] = 4 + 39 * 5
a[40] = 4 + 195
a[40] = 199
Therefore, the 40th term of the arithmetic progression is 199.
To find the 40th term of the arithmetic progression (A.P.), we can use the formula for the nth term of an arithmetic sequence:
nth term = a + (n - 1)d,
Where:
- nth term is the term we want to find,
- a is the first term of the sequence,
- n is the position of the term we want to find, and
- d is the common difference between consecutive terms.
First, we can find the common difference (d) using the given information. We know that the 7th term is 34 and the 15th term is 74.
Using the formula for the nth term, we can substitute the values of a, n, and the known 7th term (34):
34 = a + (7 - 1)d
Simplifying, we get:
34 = a + 6d ----(1)
Similarly, using the formula for the nth term, we substitute the values of a, n, and the known 15th term (74):
74 = a + (15 - 1)d
Simplifying, we get:
74 = a + 14d ----(2)
Now we have a system of two equations (equations (1) and (2)) with two variables (a and d). We can solve this system of equations to find the values of a and d.
Subtracting equation (1) from equation (2), we eliminate "a":
74 - 34 = (a + 14d) - (a + 6d)
Simplifying, we get:
40 = 8d
Dividing both sides of the equation by 8, we find:
d = 5
Now that we have the value of d, we can substitute it into equation (1) to solve for a:
34 = a + 6(5)
34 = a + 30
Subtracting 30 from both sides of the equation, we find:
a = 4
We now know that the first term (a) is 4 and the common difference (d) is 5.
To find the 40th term, we can again use the nth term formula:
40th term = a + (40 - 1)d
Substituting the values we found for a and d, we have:
40th term = 4 + (40 - 1)5
Simplifying,
40th term = 4 + 39 * 5
40th term = 4 + 195
Finally,
40th term = 199.
T15 - T7 = 74-34 = 40 = 8d
so, d=5
T40 = T15 + 25d = 74 + 25*5 = ?