In an arithmetic sequence, if a2= 5 and a10= 19 find a6
1 answer:
Answer:
a₆ = 12
Step-by-step explanation:
The nth term of an arithmetic sequence is
= a₁ + (n - 1)d
where a₁ is the first term and d the common difference
Given a₂ = 5 and a₁₀ = 19 , then
a₁ + d = 5 → (1)
a₁ + 9d = 19 → (2)
Subtract (1) from (2) term by term to eliminate a₁
8d = 14 ( divide both sides by 8 )
d = 1.75
Substitute d = 1.75 into (1)
a₁ + 1.75 = 5 ( subtract 1.75 from both sides )
a₁ = 3.25
Then
a₆ = 3.25 + 5(1.75) = 3.25 + 8.75 = 12
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Answer:
390
Step-by-step explanation:
The formula of sum of arithmetic sequence is
The number of this series are 2, 4, 6, 8, 10, 12, . . . . , 60.
The first term ( a ) = 2
The 30th even number is 60
The Tn is the term of the sequence
So the equation is
30/2 x ( 2 + )
= 60
30/2 x ( 2 + 60 )
30/2 x (62)
1860/2
390
Answer:
This contradicts the Mean Value Theorem since there exists a c on (1, 7) such that f '(c) = f(7) − f(1) (7 − 1) , but f is not continuous at x = 3
Step-by-step explanation:
The given function is
When we differentiate this function with respect to x, we get;
We want to find all values of c in (1,7) such that f(7) − f(1) = f '(c)(7 − 1)
This implies that;
If this function satisfies the Mean Value Theorem, then f must be continuous on [1,7] and differentiable on (1,7).
But f is not continuous at x=3, hence this hypothesis of the Mean Value Theorem is contradicted.