# Thread: Arithmetic series..... Log.. other problems..

1. ## Arithmetic series..... Log.. other problems..

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.

???

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??

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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)

???????????

2. Originally Posted by ansonbound
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.

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

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

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

$\color{blue}a+20d=2a+30d$

subtract the left from the right

$\color{blue}0=a+10d$

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

$\color{blue}5d=10\ \Rightarrow\ d=2,\ a=-20$

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

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

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

Inside the brackets is zero, so

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

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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??

$\color{blue}log_9x=0\ \Rightarrow\ x=9^0=\frac{9^y}{9^y}=1$

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

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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

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

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

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

3. Thankyou very much!!! now i understand!!!!