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GenaCL600 [577]
2 years ago
13

Write an equation for the nth term of the arithmetic sequence. Then find a40.

Mathematics
1 answer:
Olegator [25]2 years ago
8 0

Answer:

The nth term is: a_n = \frac{1}{4} + \frac{1}{4}(n-1)

a40 = 10

Step-by-step explanation:

Arithmetic sequence:

In an arithmetic sequence, the difference between consecutive terms is always the same, and this difference is called common difference.

The nth term of a sequence is given by:

a_n = a_1 + (n-1)d

In which a_1 is the first term and d is the common difference.

1/4,1/2

This means that:

d = \frac{1}{2} - \frac{1}{4} = \frac{2}{4} - \frac{1}{4} = \frac{1}{4}

1/4

This means that a_1 = \frac{1}{4}

The nth term is:

a_n = a_1 + (n-1)d

a_n = \frac{1}{4} + \frac{1}{4}(n-1)

Then find a40.

a_{40} = \frac{1}{4} + \frac{1}{4}(40-1) = \frac{1}{4} + \frac{39}{4} = \frac{40}{4} = 10

So

a40 = 10

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2 years ago
Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:
kondaur [170]
\text{Proof by induction:}
\text{Test that the statement holds or n = 1}

LHS = (3 - 2)^{2} = 1
RHS = \frac{6 - 4}{2} = \frac{2}{2} = 1 = LHS
\text{Thus, the statement holds for the base case.}

\text{Assume the statement holds for some arbitrary term, n= k}
1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}

\text{Prove it is true for n = k + 1}
RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}

LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}
= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}
= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}
= \frac{k(6k^{2} - 3k - 1) + 2(3k + 1)^{2}}{2}
= \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}
= \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}
= \frac{k(6k^{2} + 15k + 11) + 2}{}
= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}
= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}
= RHS

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.
3 0
3 years ago
4x+y=17
faltersainse [42]

Answer:

x=4

y=1

Step-by-step explanation:

4x=7y+9

4x+y=17

7y+9+y=17

8y=17-9

8y=8

y=1

so

4x=7*1+9

4x=16

x=16/4

x=4

4 0
2 years ago
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