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3 years ago

the first four terms of an arithmetic sequence are 2,a-b, 2a+b+7 and a - 3b, where a and b are constants. Find a and b

Mathematics

1 answer:

jonny [76]3 years ago

7 0

Arithmetic Sequence

The values of a and b is a = 2 and b=3

Step-by-step explanation:

Given the terms of the arithmetic sequence are

2 , a - b , 2a + b + 7, a - 3b

Let the common difference be D

Therefore,

The difference between the first two consecutive terms is

(a – b) – 2 = D ------------------------------( 1 )

The difference between the next two consecutive terms is

D = (2a + b+7) – ( a - b ) ---------------------(2 )

Equating equation 1 and equation 2

⇒ (a – b) -2 =(2a+b+7)-(a-b)

⇒ a – b – 2 = a + 2b +7

⇒ 3b = -9

⇒ b = -3

Similarly

The difference between the next two consecutive terms is

D = (a-3b)-(2a+b+7) ------- (3)

⇒ (a-3b)-(2a+b+7)=(2a+b+7)-(a-b)

⇒ a-3b)-(2a+b+7) -a - 4b -7 === a+2b+7

⇒ 2a = - ( 14 + 6b)

⇒ a = -( 7 + 3b)

⇒ a = - ( 7 – 3*3 )

Thus the value of a = 2

Hence , the values of a and b is a = 2 and b=3

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Use the method of "undetermined coefficients" to find a particular solution of the differential equation. (The solution found ma

Naddika [18.5K]

Answer:

The particular solution of the differential equation

= $\frac{-1,36,656cos5x+1,89,800 sin5x}{-1204}$ + $\frac{1}{37}185e^{6x})$

Step-by-step explanation:

Given differential equation y''(x) − 10y'(x) + 61y(x) = −3796 cos(5x) + 185e6x

The differential operator form $(D^{2} -10D+61)y(x) = −3796 cos(5x) + 185e^{6x}$

<u>Rules for finding particular integral in some special cases:-</u>

let f(D)y = $e^{ax}$ then

the particular integral $\frac{1}{f(D)} (e^{ax} ) = \frac{1}{f(a)} (e^{ax} ) if f(a)$ ≠ 0

let f(D)y = cos (ax ) then

the particular integral $\frac{1}{f(D)} (cosax ) = \frac{1}{f(D^2)} (cosax ) =\frac{cosax}{f(-a^2)}$ f(-a^2) ≠ 0

Given problem

$(D^{2} -10D+61)y(x) = −3796 cos(5x) + 185e^{6x}$

P<u>articular integral</u>:-

$P.I = \frac{1}{f(D)}( −3796 cos(5x) + 185e^{6x})$

$P.I = \frac{1}{D^2-10D+61}( −3796 cos(5x) + 185e^{6x})$

$P.I = \frac{1}{D^2-10D+61}( −3796 cos(5x) + \frac{1}{D^2-10D+61}185e^{6x})$

P.I = $I_{1} +I_{2}$

we will apply above two conditions, we get

$I_{1}$ =

$\frac{1}{D^2-10D+61}( −3796 cos(5x) = \frac{1}{(-25)-10D+61}( −3796 cos(5x) ( since D^2 = - 5^2)$ $= \frac{1}{(36-10D}( −3796 cos(5x) \\= \frac{1}{(36-10D}X\frac{36+10D}{36+10D} ( −3796 cos(5x)$

on simplification we get

= $\frac{1}{(36^2-(10D)^2}36+10D( −3796 cos(5x)$

= $\frac{-1,36,656cos5x+1,89,800 sin5x}{1296-100(-25)}$

= $\frac{-1,36,656cos5x+1,89,800 sin5x}{-1204}$

$I_{2}$ =

$\frac{1}{D^2-10D+61}185e^{6x}) = \frac{1}{6^2-10(6)+61}185e^{6x})$

$\frac{1}{37}185e^{6x})$

Now particular solution

P.I = $I_{1} +I_{2}$

P.I = $\frac{-1,36,656cos5x+1,89,800 sin5x}{-1204}$ + $\frac{1}{37}185e^{6x})$

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Answer with Step-by-step explanation:

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