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Thread: algebra problems

  1. #1
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    Red face algebra problems

    please friends help me outwith these sums

    1)If the sum of first n terms of two arithmetic progression are in the ratio
    (7n + 1) 4n + 27), find the ratio of their 11th terms.

    2)Find the sum of first n terms of the series
    3 8 + 6 11 + 9 14 + ....

    3)Let f(x) = x2 + 3, g(x) = x 2, then find fog and gof. Also, show that (fog) (3/2) =(gof) (3/2).

    4)Solve the equation:
    Sin x + Sin 2x + Sin 3x + Sin 4x = 0
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  2. #2
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    Quote Originally Posted by rizwanrivz View Post
    please friends help me outwith these sums

    1)If the sum of first n terms of two arithmetic progression are in the ratio
    (7n + 1) 4n + 27), find the ratio of their 11th terms.
    sol1]
    let the first series be a ,a+d,a+2d,.....

    the second series be A,A+D, A+2D,.......
    given s/S=(7n+1)4n+27)

    TO FIND--------- t11/T11=? .............t11 denotes 11th term of first AP
    and T11 denotes 11th term of Second AP

    s=n/2*[2a+(n-1)d] ................................. sum of first AP upto n terms

    similarly sum of second AP is

    S=n/2*[2A+(n-1)D]

    s/S=(7n+1)4n+27)

    [n/2*{2a+(n-1)d}] / [n/2*{2A+(n-)D}]=(7n+1)4n+27)

    n/2 gets cancelled we get

    [2a+(n-1)d]/ [2A+(n-)D]=(7n+1)4n+27)

    dividing nr.& dr. of LHS by2 we get

    [a+(n-1)d/2] / [A+(n-1)D/2] = (7n+1)4n+27)

    now put n=21 both sides we get

    [a+10d]/[A+10D]=148/111

    t11/T11=148/111
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  3. #3
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    Quote Originally Posted by rizwanrivz View Post
    please friends help me outwith these sums

    1)If the sum of first n terms of two arithmetic progression are in the ratio
    (7n + 1) 4n + 27), find the ratio of their 11th terms.
    Sum of n terms of an A.P.
    $\displaystyle =\frac{n}{2} * (2a+(n-1)d)$
    Put $\displaystyle a_1 $and $\displaystyle a_2 $for two APs and d1 and d2 respectively as the common difference
    so
    $\displaystyle \frac{S_1}{S_2}=\frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2} $
    Here take out 2 as common and cancel it ie;

    $\displaystyle \frac{a_1+\frac{(n-1)d_1}{2}}{a_2+\frac{(n-1)d_2}{2}}
    $
    Now here put the value of $\displaystyle \frac{n-1}{2}=11$this will give you the value of n
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  4. #4
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    Quote Originally Posted by rizwanrivz View Post
    please friends help me outwith these sums


    2)Find the sum of first n terms of the series
    3 8 + 6 11 + 9 14 + ....
    the general term of the given series is Tn=3n*(3n+5)

    Tn=9n^2+15n

    now taking summation both sides
    ΣTn=9*Σn^2 + 15*Σn................................(1)

    now we know that
    Σn=n(n+1)/2
    and Σn^2=n(n+1)(2n+1)/6
    put this in eq. (1) and get the answer
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  5. #5
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    Quote Originally Posted by rizwanrivz View Post
    please friends help me outwith these sums



    3)Let f(x) = x2 + 3, g(x) = x 2, then find fog and gof. Also, show that (fog) (3/2) =(gof) (3/2).
    fog(x)= (x-2)^2+3
    and
    gof(x)=x^2+3-2

    Put 3/2 in both of them and check yourself
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  6. #6
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    Hello, rizwanrivz!

    For #4, we need two sum-to-product identities:

    . . $\displaystyle \sin A + \sin B \:=\:2\sin\left(\tfrac{A+B}{2}\right)\cos\left(\tf rac{A-B}{2}\right)$

    . . $\displaystyle \cos A + \cos B \:=\:2\cos\left(\tfrac{A+B}{2}\right)\cos\left(\tf rac{A-B}{2}\right)$


    4) Solve the equation: .$\displaystyle \sin x + \sin 2x + \sin 3x + \sin 4x \:= \:0$

    $\displaystyle \text{We have: }\;\underbrace{\sin x + \sin4x} + \underbrace{\sin2x + \sin3x} \:=\:0$

    . . . . $\displaystyle \overbrace{2\sin\left(\tfrac{5x}{2}\right)\cos\lef t(\tfrac{3x}{2}\right)} + \overbrace{2\sin\left(\tfrac{5x}{2}\right)\cos\lef t(\tfrac{7x}{2}\right)} \:=\:0 $

    Factor:. . $\displaystyle 2\sin\left(\tfrac{5x}{2}\right)\underbrace{\bigg[\cos\left(\tfrac{3x}{2}\right) + \cos\left(\tfrac{7x}{2}\right)\bigg]} \:=\:0 $

    . . . . . . . . $\displaystyle 2\sin\left(\tfrac{5x}{2}\right)\cdot \overbrace{2\cos(5x)\cos(2x)} \:=\:0 $

    . . . . . . . . . $\displaystyle 4\sin\left(\tfrac{5x}{2}\right)\cdot\cos5x\cdot \cos2x \:=\:0$


    Then: .$\displaystyle \begin{array}{ccccccc}

    \sin\frac{5x}{2} \:=\:0 & \rightarrow & \frac{5x}{2} \:=\:\pi n & \Rightarrow & x \:=\:\frac{2\pi}{5}n \\
    \cos5x \:=\:0 & \Rightarrow & 5x \:=\:\frac{\pi}{2} + \pi n & \Rightarrow & x \:=\:\frac{\pi}{10} + \frac{\pi}{5}n \\ \\[-4mm]
    \cos2x \:=\:0 & \Rightarrow & 2x \:=\:\frac{\pi}{2} + \pi n & \Rightarrow & x \:=\:\frac{\pi}{4} + \frac{\pi}{2}n

    \end{array}$

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