**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.

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

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

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)

???????????

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
$\displaystyle \color{blue}y-49=-\frac{1}{45}(x-1)$