# The even solution of equation dy/dx=12 is the first term of an arithmetical sequence with common difference 4. Find a7 and a11. y=2x^3-3x^2

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We have y = 2x^3 - 3x^2.

The solution of dy/dx = 12 is the first term of the arithmetic sequence with common difference 4.

dy/dx = 12

=> 6x^2 - 6x = 12

=> x^2 - x = 2

=> x^2 - 2x + x - 2 = 0

=> x(x - 2) + 1(x - 2) = 0

=> (x + 1)(x - 2) = 0

The even root is 2.

The nth term of an arithmetic sequence is a + (n - 1)d

a = 2 , d = 4

a7 = 2 + 6*4 = 26

a11 = 2 + 10*4 = 42

**The required terms: a7 = 26 and a11 = 42**

To determine the 7th and the 11th terms of the arithmetical sequence, we'll have to determine the 1st term.

From enunciation, we know that the first term of the arithmetical sequence is also the solution of the equatin dy/dx = 12.

We'll differentiate the given y with respect to x.

dy/dx = d(2x^3-3x^2)/dx

dy/dx = 6x^2 - 6x

We'll re-write the equation:

6x^2 - 6x = 12

We'll divide by 6:

x^2 - x = 2

We'll subtract 2:

x^2 - x - 2 = 0

We'll apply quadratic formula:

x1 = [1 + sqrt(1 + 8)]/2

x1 = (1+3)/2

x1 = 2

x2 = (1-3)/2

x2 = -1

The even solution is x = 2.

The first term of the arithmetical sequence is a1 = 2.

We'll write the general term of an arithmetical sequence:

an = a1 + (n-1)*d, where d is the common difference

a7 = a1 + 6d

a7 = 2 + 6*4

a7 = 2 + 24

a7 = 26

a11 =a1 + 10d

a11 = 2 + 10*4

a11 = 2 + 40

a11 = 42

**The 7th and the 11th terms of the arithmetical sequence are: **

**a7 = 26 ; a11 = 42.**