1. ## 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

2. Originally Posted by rizwanrivz
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

3. Originally Posted by rizwanrivz
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.
$=\frac{n}{2} * (2a+(n-1)d)$
Put $a_1$and $a_2$for two APs and d1 and d2 respectively as the common difference
so
$\frac{S_1}{S_2}=\frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}$
Here take out 2 as common and cancel it ie;

$\frac{a_1+\frac{(n-1)d_1}{2}}{a_2+\frac{(n-1)d_2}{2}}
$

Now here put the value of $\frac{n-1}{2}=11$this will give you the value of n

4. Originally Posted by rizwanrivz
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

5. Originally Posted by rizwanrivz
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

6. Hello, rizwanrivz!

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

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

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

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

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

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

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

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

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

Then: . $\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}$