The fourth term of an AP is 11 and the eighth term exceeds twice the fourth term by five.Find the AP and the sum of first 50 terms.
a1 = the initial term of an arithmetic progression
d = the common difference of successive members
nth term of the sequence:
an = a1 + ( n - 1 ) ∙ d
In this case:
a4 = 11
a4 = a1 + ( 4 - 1 ) ∙ d
a4 = a1 + 3 d
11 = a1 + 3 d
a8 = 2 a4 + 5
a8 = 2 ∙ 11 + 5 = 22 + 5 = 27
a8 = a1 + ( 8 - 1 ) ∙ d
a8 = a1 + 7 d
a8 - a 4 = a1 + 7 d - ( a1 + 3 d ) =
a1 + 7 d - a1 - 3 d = 7 d - 3 d = 4 d
a8 - a 4 = 4 d = 27 - 11 = 16
4 d = 16
d = 16 / 4 = 4
d = 4
a4 = a1 + 3 d
11 = a1 + 3 ∙ 4
11 = a1 + 12
11 - 12 = a1
- 1 = a1
a1 = - 1
The sum of the n members of a arithmetic progression:
Sn = ( n / 2 ) ∙ [ 2 a1+ ( n - 1 ) ∙ d ]
In this case:
S50 = ( 50 / 2 ) ∙ [ 2 ∙ ( - 1 ) + ( 50 - 1 ) ∙ 4 ]
S50 = 25 ∙ ( - 2 + 49 ∙ 4 )
S50 = 25 ∙ ( - 2 + 196 )
S50 = 25 ∙ 194
S50 = 4850
Let's use the general formula for the nth term of an arithmetic progression (AP):
an = a1 + (n-1)d
where:
an = nth term of the AP
a1 = first term of the AP
d = common difference of the AP
Given that the fourth term (a4) is 11, we can substitute n = 4 and an = 11 into the formula:
11 = a1 + (4-1)d
11 = a1 + 3d
We are also given that the eighth term (a8) exceeds twice the fourth term by five. In other words, a8 = 2(a4) + 5 = 2(11) + 5 = 27. Substituting n = 8 and an = 27 into the formula:
27 = a1 + (8-1)d
27 = a1 + 7d
Now we have a system of two equations:
11 = a1 + 3d ----(1)
27 = a1 + 7d ----(2)
We can solve this system of equations to find the values of a1 and d.
Subtracting equation (1) from equation (2), we get:
27 - 11 = (a1 + 7d) - (a1 + 3d)
16 = 4d
d = 4
Substituting d = 4 back into equation (1), we can find a1:
11 = a1 + 3(4)
11 = a1 + 12
a1 = -1
Therefore, the first term (a1) of the arithmetic progression is -1 and the common difference (d) is 4.
Now we can find the AP by substituting these values into the formula:
an = a1 + (n-1)d
Let's find the first 50 terms of the AP and calculate their sum:
Sum = (number of terms / 2) * (first term + last term)
The first term (a1) is -1, and we can find the 50th term (a50) using the formula:
a50 = a1 + (50-1)d
a50 = -1 + 49(4)
a50 = -1 + 196
a50 = 195
Substituting these values into the sum formula:
Sum = (50 / 2) * (-1 + 195)
Sum = 25 * 194
Sum = 4850
Therefore, the sum of the first 50 terms of the arithmetic progression is 4850.
To find the terms of an arithmetic progression (AP) and the sum of its terms, we can use the formulas associated with APs.
Let's say the first term of the AP is 'a', and the common difference is 'd'.
Given that the fourth term is 11, we know that:
a + (4-1)d = 11 ----(1)
We can simplify this equation to:
a + 3d = 11 ----(2)
Now, we are also given that the eighth term exceeds twice the fourth term by five. So, the eighth term is (2*11+5) = 27.
Using the formula for the nth term of an AP, we can write this as:
a + (8-1)d = 27 ----(3)
Simplifying this equation gives:
a + 7d = 27 ----(4)
We now have a system of equations (2) and (4) with two variables 'a' and 'd'. We can solve this system to find the values of 'a' and 'd'.
Subtracting equation (2) from equation (4), we get:
a + 7d - (a + 3d) = 27 - 11
Simplifying, we have:
a + 4d = 16 ----(5)
Solving equations (2) and (5) simultaneously will give us the values of 'a' and 'd'.
Subtracting equation (2) from equation (5), we get:
a + 4d - (a + 3d) = 16 - 11
Simplifying, we have:
d = 5 ----(6)
Substituting this value of 'd' into equation (2), we get:
a + 3(5) = 11
a + 15 = 11
a = -4
Therefore, the first term 'a' is -4 and the common difference 'd' is 5.
Hence, the AP is: -4, 1, 6, 11, 16, 21, 26, 31, ...
To find the sum of the first 50 terms, we can use the formula for the sum of an AP:
Sum (Sn) = (n/2) * [2a + (n-1)d]
Substituting the values of 'a', 'd', and 'n' (n = 50) into this formula, we can calculate the sum:
Sum (S50) = (50/2) * [2(-4) + (50-1)5]
Sum (S50) = 25 * [-8 + 49 * 5]
Sum (S50) = 25 * [-8 + 245]
Sum (S50) = 25 * 237
Sum (S50) = 5925
Therefore, the sum of the first 50 terms of the given AP is 5925.