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stira [4]
4 years ago
10

Find S18 for geometric series given a5=-6 and a2= -48

Mathematics
1 answer:
Ganezh [65]4 years ago
3 0

an = a1r^(n-1)

a5 = a1 r^(5-1)

-6 =a1 r^4


a2 = a1 r^(2-1)

-48 = a1 r


divide

-6 =a1 r^4

----------------    yields   1/8 = r^3      take the cube root  or each side

-48 = a1 r                     1/2 = r


an = a1r^(n-1)

an = a1 (1/2)^ (n-1)

-48 = a1 (1/2) ^1

divide by 1/2

-96 = a1


an = -96 (1/2)^ (n-1)


the sum

Sn = a1[(r^n - 1/(r - 1)]

S18 = -96 [( (1/2) ^17 -1/ (1/2 -1)]

       =-96 [ (1/2) ^ 17 -1 /-1/2]

      = 192 * [-131071/131072]

 approximately -192

     

       

     

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Evaluate the line integral, where c is the given curve. C xeyz ds, c is the line segment from (0, 0, 0) to (2, 3, 4)
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The value of line integral is, 73038 if the c is the given curve. C xeyz ds, c is the line segment from (0, 0, 0) to (2, 3, 4)

<h3>What is integration?</h3>

It is defined as the mathematical calculation by which we can sum up all the smaller parts into a unit.

The parametric equations for the line segment from (0, 0, 0) to (2, 3, 4)

x(t) = (1-t)0 + t×2 = 2t  

y(t) = (1-t)0 + t×3 = 3t

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Finding its derivative;

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The line integral is given by:

\rm \int\limits_C {xe^{yz}} \, ds = \int\limits^1_0 {2te^{12t^2}} \, \sqrt{2^2+3^2+4^2} dt

 

\rm ds = \sqrt{2^2+3^2+4^2} dt

After solving the integration over the limit 0 to 1, we will get;

\rm \int\limits_C {xe^{yz}} \, ds = \dfrac{\sqrt{29}}{12}  (e^{12}-1)   or

= 73037.99 ≈ 73038

Thus, the value of line integral is, 73038 if the c is the given curve. C xeyz ds, c is the line segment from (0, 0, 0) to (2, 3, 4)

Learn more about integration here:

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