Solve the following problems. 1. Find the arithmetic mean between-14 and -30 2. Insert two anithmetic means between x + y and 4x - 2y. 3 The arithmetic menn between two terms in an arithmetic sequence is 49. if one of these terms is 42 find the other term. 4 If five antimetic means are inserted between -9 and 9, what is the third anthmetic mean?
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Answer:
1.60.
2.2x and 3x-y.
3. Xₙ = a + d(n-1)
Step-by-step explanation:
1. The arithmetic mean between a and 10 is 30, the value of 'a' should be. 45.
2. asubn=asub1 + (n-1)d
asub4=asub1 + (n-1)d
4x - 2y=x + y + (4-1)d
4x - x - 2y - y = 3d
3x - 3y=3d
d=3(x-y)/3
d= x-y
3.Arithmetic Sequence:
An Arithmetic sequence is a series of number in order where the difference between one term and the next is a constant.
It follows the rule below:
Xₙ = a + d(n-1)
Where;
a = is the first term
d = common difference
Arithmetic Mean
It is the average value among a set of numbers
It is given by the formula:
Mean = ∑¹ₙ\frac{Xi + Xn}{n}
n
Xi+Xn
where;
x = the sequence of terms
n = the number of terms
To solve the given problem,
Given:
Mean = 39
X₂ = 32
n = 2 → number of terms
Required:
X₁ = First term
Solution:
Using Arithmetic Mean:
Mean = \frac{Xi + Xn}{n}
n
Xi+Xn
39 = \frac{X1 + 32}{2}
2
X1+32
2(39) = X₁+32
78 = X₁ + 32
X₁ = 78 - 32
X₁ = 46
Checking
39 = \frac{46 + 32}{2}
2
46+32
39 = 39
Thus , the other term is 46
Using Arithmetic Sequence:
Xₙ = a + d(n-1)
d = 39 - 32 considering the first and second term as we have three term if using arithmetic sequence. 32, 39, X₃
= 32 + (7) (3-1)
=32 + (7) (2)
= 32 + 14
X₃ = 46