Arithmetic series..... Log.. other problems..

Nov 2009
28
0
An arithmetic series has five term a and common difference d.
The sum of the first 31 terms of the series is 310.


a) Show that a + 15d = 10

> S31 = 31/2 ( 2a + (31-1) d)
> 31( a + 15d ) = 310
> a + 15d = 310/31; a + 15d = 10
(Rock)

b)Given also that the 21st term is twise the 16th term, find the value of d.

???

c)the nth term of the series is un. Given that
K
Sum un = 0, find the value of K.
n = 1

???

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Find the value of x in each of the following:

(A). log9x = 0
(B). log9x = 1/2

of coz i noe how to do it on the calculator with the answers 1 for (A) and 3 for (B), but how do i get it withour a calculator??

----------------------------------

A curve has equation y = 1/x^3 + 48x.

a) Find dy/dx.

>> dy/dx = 3x^-4 + 48

b) Hence find the equation of each of the two tangents to the curve that are parallel to the x-axis.

>> -3x^-4 + 48 = 0
>> x^-4 = 16
>> x = +-1/2


c) Find the equation of the normal to the curve at the point (1,49)

???????????(Angry)
 
Dec 2009
3,120
1,342
An arithmetic series has five term a and common difference d.
The sum of the first 31 terms of the series is 310.


a) Show that a + 15d = 10

> S31 = 31/2 ( 2a + (31-1) d)
> 31( a + 15d ) = 310
> a + 15d = 310/31; a + 15d = 10
(Rock)

b)Given also that the 21st term is twise the 16th term, find the value of d.

??? \(\displaystyle \color{blue}a+15d=10......clue1\)

\(\displaystyle \color{blue}a+(21-1)d=2[a+(16-1)d].......clue2\)

The clues allow you to solve the simultaneous equations for "d"

\(\displaystyle \color{blue}a+20d=2a+30d\)

subtract the left from the right

\(\displaystyle \color{blue}0=a+10d\)

use your first result....clue1

\(\displaystyle \color{blue}a+15d-(a+10d)=10-0\)

\(\displaystyle \color{blue}5d=10\ \Rightarrow\ d=2,\ a=-20\)




c)the nth term of the series is un. Given that
K
Sum un = 0, find the value of K.
n = 1

??? \(\displaystyle \color{blue}\frac{k}{2}[2a+(k-1)d]=0\)

\(\displaystyle \color{blue}\frac{k}{2}[-40+(k-1)2]=0\)

Inside the brackets is zero, so

\(\displaystyle \color{blue}(k-1)2=40\ \Rightarrow\ k-1=20\ \Rightarrow\ k=21\)


----------------------------------
Find the value of x in each of the following:

(A). log9x = 0
(B). log9x = 1/2

of coz i noe how to do it on the calculator with the answers 1 for (A) and 3 for (B), but how do i get it withour a calculator??

\(\displaystyle \color{blue}log_9x=0\ \Rightarrow\ x=9^0=\frac{9^y}{9^y}=1\)

\(\displaystyle \color{blue}log_9x=\frac{1}{2}\ \Rightarrow\ x=9^{\frac{1}{2}}=\sqrt{9}=3\)

----------------------------------

A curve has equation y = 1/x^3 + 48x.

a) Find dy/dx.

>> dy/dx = -3x^-4 + 48

b) Hence find the equation of each of the two tangents to the curve that are parallel to the x-axis.

>> -3x^-4 + 48 = 0
>> x^-4 = 16
>> x = +-1/2


c) Find the equation of the normal to the curve at the point (1,49)

???????????(Angry)

The slope of the tangent at x=1 is

\(\displaystyle \color{blue}-3(1)^{-4}+48=45\)

The normal is perpendicular to the tangent, hence invert and negate the tangent slope


The slope of the normal is -1/45 and the normal contains the point (1,49)

Hence, the normal equation at that point is

\(\displaystyle \color{blue}y-49=-\frac{1}{45}(x-1)\)
.
 
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Nov 2009
28
0
Thankyou very much!!! now i understand!!!!(Bow)(Bow)(Bow)
(Rofl)