word how there's a difficulty-loose distinction of four between consecutive words. the 1st term is 3. The formula for an arithmetic sequence is a_n = a_1 + (n - a million) d, the place a_1 is the 1st term, n is the form of the term, and d is the difficulty-loose distinction. right here a_1 = 3 and d = 4. a_n = 3 + (n - a million)(4) = 3 + 4n - 4 = 4n - a million
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There will be four roots.
Because there are no x or x^3 terms, they will be -p, -q, +q, +p.
If these are to be in arithmetic sequence then p - q = q - (-q) ----> p = 3q
Using the usual rules about product of roots of equations
-3q*-q*q*3q = (1/200)*sin(pi/6) = 1/400 ----> q^4 = 1/3600 ----> q = sqrt(1/60)
The x^2 term will then be
(-3q*-q) + (-3q*q) + (-3q*3q) + (-q*q) + (-q*3q) + (q*3q) = -10q^2 = -10*(1/60) = -1/6
Therefore (1/3)*sinB = -1/6 ----> sinB = -1/2 ----> B = 5pi/6
word how there's a difficulty-loose distinction of four between consecutive words. the 1st term is 3. The formula for an arithmetic sequence is a_n = a_1 + (n - a million) d, the place a_1 is the 1st term, n is the form of the term, and d is the difficulty-loose distinction. right here a_1 = 3 and d = 4. a_n = 3 + (n - a million)(4) = 3 + 4n - 4 = 4n - a million
let be
x1=a
x2=a+r
x3 =a+2r
x4=a+3r
roots of your equation where r ratio and a a real number
then sum of them
x1+x2+x3+x4= 0 (coefficient of x^3 it's 0)
x1+x2+x3+x4 = 4a+6r =0 then
r = -2a/3
Now roots are
x1=a
x2=a -2a/3 = a/3
x3 =a -4a/3 = -a/3
x4=a- 6a/3 = - a
but product of all roots is :
a*a/3*(-a)/3*(-a) = a^4/9 = 1/400 (fre term of your equation)
then a^4 = 3600
a^2 = 60 or -60 but no permitted
then a = sqrt(60) or a = -sqrt(60)
Then roots are :
x1 = sqrt(60)
x2 = sqrt(60)/3
x3 = - sqrt(60)/3
x4 =- sqrt(60)
then (1/3) sin(b ) = x1x2+x1x3+x1x4+x2x3+x2x4+x3x4 =.....