An arithmetic sequence has a1 = 22 and a8 = 43. Find a15.

Mathematics

1 answer:

brilliants [131]2 years ago

4 0

The 15th term of the arithmetic sequence $a_{15} = 64$

Option B is correct

The nth term of an arithmetic sequence is given as:

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

The first value, a = 22

$a_8 = a+(8 - 1)d\\a_8 = a + 7d$

Since $a_8 = 43$

$43 = 22 + 7d\\7d = 43 - 22\\7d = 21\\d = \frac{21}{7}\\d = 3$

The common difference, d = 3

$a_{15}= a + (15-1)d\\a_{15} = a + 14d\\a_{15} = 22 + 14(3)\\a_{15} = 22 + 42\\a_{15} = 64$

The 15th term of the arithmetic sequence = 64

Learn more here: brainly.com/question/24072079

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Please help me with this question asap

Elden [556K]

3 because if the graph shows 1,2 there should be 3 even 1+2=3

i hope that helped you if you dont get anything just ask me thank you

4 0

3 years ago

Apply Gaussian quadrature with n = 4 to approximate integrate sin x^2dx from 1 to 5

maks197457 [2]

First, recall that Gaussian quadrature is based around integrating a function over the interval [-1,1], so transform the function argument accordingly to change the integral over [1,5] to an equivalent one over [-1,1].

$x=2t+3\iff t=\dfrac x2-\dfrac32\implies2\mathrm dt=\mathrm dx$
$x=1\implies t=\dfrac{2-6}4=-1$
$x=5\implies t=\dfrac{10-6}4=1$

So,

$\displaystyle\int_{x=1}^{x=5}\sin x^2\,\mathrm dx=\displaystyle2\int_{t=-1}^{t=1}\sin(2t+3)^2\,\mathrm dt$

Let $f(t)=2\sin(2t+3)^2$ . With $n=4$ , we're looking for coefficients $c_i$ and nodes $x_i$ , with $1\le i\le4$ , such that

$\displaystyle\int_{-1}^1f(t)\,\mathrm dt\approx c_1f(x_1)+\cdots+c_4f(x_4)$

You can either try solving for each with the help of a calculator, or look up the values of the weights and nodes (they're extensively tabulated, and I'll include a link to one such reference).

Using the quadrature, we then have

$\displaystyle\int_{-1}^1f(t)\,\mathrm dt\approx0.3749f(-0.8611)+0.6521f(-0.3400)+0.6521f(0.3400)+0.3749f-0.8611)$
$\displaystyle\int_{-1}^1f(t)\,\mathrm dt\approx0.5790$