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Aleonysh [2.5K]
3 years ago
8

If 2k, 5k-1 and 6k+2 are the first 3 terms of an arithmetic sequence, find k and the 8th term.

Mathematics
1 answer:
Simora [160]3 years ago
4 0
Consecutive terms in an arithmetic sequence differ by a constant d. So

5k-1=2k+d
6k+2=5k-1+d
\implies\begin{cases}3k-d=1\\k-d=-3\end{cases}\implies k=2,d=5

Denote the n-th term in the sequence by a_n. Now that a_1=2, we have

a_n=a_{n-1}+5=a_{n-2}+2\cdot5=\cdots=a_1+(n-1)\cdot5

which means

a_8=2+(8-1)\cdot5=37
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Step-by-step explanation:

Question:-

- We are given the following non-homogeneous ODE as follows:

                           x^2y' +xy = 3

- A general solution to the above ODE is also given as:

                          y = \frac{3Ln(x) + C  }{x}

- We are to prove that every member of the family of curves defined by the above given function ( y ) is indeed a solution to the given ODE.

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- To determine the validity of the solution we will first compute the first derivative of the given function ( y ) as follows. Apply the quotient rule.

                          y' = \frac{\frac{d}{dx}( 3Ln(x) + C ) . x - ( 3Ln(x) + C ) . \frac{d}{dx} (x)  }{x^2} \\\\y' = \frac{\frac{3}{x}.x - ( 3Ln(x) + C ).(1)}{x^2} \\\\y' = - \frac{3Ln(x) + C - 3}{x^2}

- Now we will plug in the evaluated first derivative ( y' ) and function ( y ) into the given ODE and prove that right hand side is equal to the left hand side of the equality as follows:

                          -\frac{3Ln(x) + C - 3}{x^2}.x^2 + \frac{3Ln(x) + C}{x}.x = 3\\\\-3Ln(x) - C + 3 + 3Ln(x) + C= 3\\\\3 = 3

- The equality holds true for all values of " C "; hence, the function ( y ) is the general solution to the given ODE.

- To determine the complete solution subjected to the initial conditions y (1) = 3. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

                         y( 1 ) = \frac{3Ln(1) + C }{1} = 3\\\\0 + C = 3, C = 3

- Therefore, the complete solution to the given ODE can be expressed as:

                        y ( x ) = \frac{3Ln(x) + 3 }{x}

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                        y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{y}

                           

Download docx
6 0
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