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
b)Given also that the 21st term is twise the 16th term, find the value of d.
???
![\color{blue}a+(21-1)d=2[a+(16-1)d].......clue2](http://latex.codecogs.com/png.latex?\color{blue}a+(21-1)d=2[a+(16-1)d].......clue2)
The clues allow you to solve the simultaneous equations for "d" 
subtract the left from the right 
use your first result....clue1 =10-0)
c)the nth term of the series is un. Given that
K
Sum un = 0, find the value of K.
n = 1
???
![\color{blue}\frac{k}{2}[2a+(k-1)d]=0](http://latex.codecogs.com/png.latex?\color{blue}\frac{k}{2}[2a+(k-1)d]=0)
![\color{blue}\frac{k}{2}[-40+(k-1)2]=0](http://latex.codecogs.com/png.latex?\color{blue}\frac{k}{2}[-40+(k-1)2]=0)
Inside the brackets is zero, so 2=40\ \Rightarrow\ k-1=20\ \Rightarrow\ k=21)
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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??
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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)
???????????
The slope of the tangent at x=1 is ^{-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 )
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Originally Posted by ansonbound 
