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2 years ago

Generate the first 5 terms of the sequence 5n - 1?

Mathematics

1 answer:

GrogVix [38]2 years ago

8 0

Answer:

The first five terms of the sequence

4 , 9 , 14 , 19 , 24

Step-by-step explanation:

Explanation:-

Given $n^{th}$ of the sequence

aₙ = 5 n -1

Put n=1

a₁ = 5(1)-1 = 5-1 =4

First term of the sequence = 4

Put n =2

a₂ = 5(2) -1 = 10-1 =9

Second term of the sequence = 9

Put n=3

a₃ = 5(3)-1 =15-1 =14

Third term of the sequence = 14

Put n=4

a₄ = 5(4)-1 = 20 -1 = 19

Fourth term of the sequence = 19

Put n=5

a₅ = 5(5)-1 = 25-1 =24

Fifth term of the sequence = 24

Final answer:-

The first five terms of the sequence

4 , 9 , 14 , 19 , 24 this is an arithmetic progression

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Read 2 more answers

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leonid [27]

Answer with explanation:

The given differential equation is

y" -y'+y=2 sin 3x------(1)

Let, y'=z

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$\frac{dy}{dx}=z\\\\d y=zdx\\\\y=z x$

Substituting the value of , y, y' and y" in equation (1)

z'-z+zx=2 sin 3 x

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This is a type of linear differential equation.

Integrating factor

$=e^{\int (x-1) dx}\\\\=e^{\frac{x^2}{2}-x}$

Multiplying both sides of equation (1) by integrating factor and integrating we get

$\rightarrow z\times e^{\frac{x^2}{2}-x}=\int 2 sin 3 x \times e^{\frac{x^2}{2}-x} dx=I$

$I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx -\int \frac{2\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx-\frac{2I}{3}\\\\\frac{5I}{3}=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{5}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{5} dx$