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3 years ago

The 9th term of an arithmetic series is 5/2

Mathematics

1 answer:

Verdich [7]3 years ago

8 0

Answer:

- 8050

Step-by-step explanation:

The n th term of an arithmetic sequence is

$a_{n}$ = a + (n - 1)d

where a is the first term and d the common difference.

We require to find both a and d

Given the 9 th term is 2.5 , then

a + 8d = 2.5 → (1)

Given the sum of the second and fifth term is 27, then

a + d + a + 4d = 27, that is

2a + 5d = 27 → (2)

Multiply (1) by - 2 and add to (2) to eliminate a

- 2a - 16d = - 5 → (3)

Add (2) and (3) term by term

- 11d = 22 ( divide both sides by - 11 )

d = - 2

Substitute d = - 2 into (1) and solve for a

a - 16 = 2.5 ( add 16 to both sides )

a = 18.5

The sum to n terms of an arithmetic sequence is

$S_{n}$ = $\frac{n}{2}$ [ 2a + (n - 1)d ], thus

$S_{100}$ = 50 [ (2 × 18.5) + (99 × - 2) ] = 50(37 - 198) = 50(- 161) = - 8050

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3 years ago

Give 3 pairs of prime numbers that add up to 50

aleksley [76]

Answer:

3 and 47

7 and 43

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Step-by-step explanation:

Prime numbers up to 50:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47

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3 years ago

Evaluate the indefinite integral. <br> integar x4/1 + x^10 dx

ivann1987 [24]

Answer:

$\int\ {\frac{x^4}{1 + x^{10}}} \, dx = \frac{1}{5}( \arctan(x^5)) + c$

Step-by-step explanation:

Given

$\int\ {\frac{x^4}{1 + x^{10}}} \, dx$

Required

Integrate

We have:

$\int\ {\frac{x^4}{1 + x^{10}}} \, dx$

Let

$u = x^5$

Differentiate

$\frac{du}{dx} = 5x^4$

Make dx the subject

$dx = \frac{du}{5x^4}$

So, we have:

$\int\ {\frac{x^4}{1 + x^{10}}} \, dx$

$\int\ {\frac{x^4}{1 + x^{10}}} \, \frac{du}{5x^4}$

$\frac{1}{5} \int\ {\frac{1}{1 + x^{10}}} \, du$

Express x^(10) as x^(5*2)

$\frac{1}{5} \int\ {\frac{1}{1 + x^{5*2}}} \, du$

Rewrite as:

$\frac{1}{5} \int\ {\frac{1}{1 + x^{5)^2}}} \, du$

Recall that: $u = x^5$

$\frac{1}{5} \int\ {\frac{1}{1 + u^2}}} \, du$

Integrate

$\frac{1}{5} * \arctan(u) + c$

Substitute: $u = x^5$

$\frac{1}{5} * \arctan(x^5) + c$

Hence:

$\int\ {\frac{x^4}{1 + x^{10}}} \, dx = \frac{1}{5}( \arctan(x^5)) + c$

7 0

3 years ago

Add the equation u-v/8v + 6u-3v/8v

notka56 [123]

Answer:

7u-4v/8

Step-by-step explanation:

u+6u=7u

-v+(-3v)=-v-3v=-4v

7u-4u/8

8 0

3 years ago