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3 years ago

the common ratio of a geometric series is 3 and the sum of the first 8 terms is 3280. what is the first term of the series?

Mathematics

1 answer:

andrew11 [14]3 years ago

4 0

Answer: 1

Step-by-step explanation:

The formula for calculating the sum of a Geometric series if the common ratio is greater than 1 is given as :

$S_{n}$ = $\frac{a(r^{n}-1) }{r-1}$

Where $S_{n}$ is the sum of terms , a is the first term , r is the common ratio and n is the number of terms.

From the question:

$S_{n}$ = 3280

a = ?

r = 3

n = 8

Substituting this into the formula , we have

3280 = $\frac{a(3^{8}-1) }{3-1}$

3280 = $\frac{a(6561-1)}{2}$

Multiply through by 2 , we have

6560 = a ( 6560)

divide through by 6560, we have

6560/6560 = a(6560)/6560

Therefore : a = 1

The first term of the series is thus 1

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kifflom [539]

Looks like we have

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$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(x^2+y^2+z^2)\,\mathrm dV$

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$\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\varphi\end{cases}\implies\mathrm dV=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$

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$\displaystyle\iiint_R(x^2+y^2+z^2)\,\mathrm dV=\int_0^{\pi/2}\int_0^{2\pi}\int_0^1\rho^4\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=\frac{2\pi}5$

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$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^1\left(\frac{u^3}3\sin^3v\,\vec\jmath+u^2\sin^2v\,\vec k\right)\times(-u\,\vec k)\,\mathrm du\,\mathrm dv$