1. In an arithmetic sequence, the fifth term, u5, is greater than the first term, u1. The difference between these terms is 36.

a) find the common difference
9?

The tenth term of the sequence is double the seventh term .
b) find u1
2.5?

2. Consider the function f(x) = ax^3 -3x+5, where a does not equal zero.
a) find f'(x).
3ax^2-3?
b) write down the value of f'(0).
-3?

The function has a local maximum at x=-2
c) calculate the value of a
0.25??

a) Why did the arithmetic sequence go to therapy? It had a common difference. The common difference is 9!

b) Why was the tenth term of the arithmetic sequence such a good salesperson? It doubled the seventh term! So, the first term, u1, is 2.5.

2. a) The derivative of f(x) is 3ax^2 - 3. So, f'(x) = 3ax^2 - 3.

b) If f'(0) is -3, then why is it always getting lost? Because it's negative! So, f'(0) = -3.

c) Why did the value of a feel so lucky? Because it hit the jackpot - it's 0.25! Way to go, a!

a) To find the common difference, we can use the formula for the nth term of an arithmetic sequence:

un = u1 + (n-1)d

Given that the fifth term (u5) is greater than the first term (u1) by 36, we can write the equation as:

u5 = u1 + 4d
u1 + 36 = u1 + 4d

Simplifying this equation, we find:
36 = 4d

Dividing both sides by 4, we get:
d = 9

Therefore, the common difference is 9.

b) To find u1, we can substitute the common difference (d = 9) into the equation:

u5 = u1 + 4d

Since the tenth term is double the seventh term, we have:
u10 = 2u7

Using the formula for the nth term, we can write these equations as:

u1 + 4d = 2(u1 + 6d)
u1 + 36 = 2u1 + 12d

Substituting the value of d from part a), we get:
u1 + 36 = 2u1 + 108

Rearranging this equation, we find:
u1 = -72

Therefore, u1 is equal to -72.

b) To find f'(x), we can take the derivative of the function f(x) with respect to x:

f'(x) = 3ax^2 - 3

b) To evaluate f'(0), we substitute x = 0 into the derivative:

f'(0) = 3a(0)^2 - 3
f'(0) = 0 - 3
f'(0) = -3

Therefore, the value of f'(0) is -3.

c) To find the value of a when the function has a local maximum at x = -2, we can set the derivative equal to zero:

f'(x) = 3ax^2 - 3 = 0

Simplifying this equation, we find:
3ax^2 = 3
ax^2 = 1
a = 1/x^2

Substituting x = -2, we get:
a = 1/(-2)^2
a = 1/4

Therefore, the value of a is equal to 0.25.

a) To find the common difference in an arithmetic sequence, we need to find the difference between any two consecutive terms. In this case, the difference between the fifth term (u5) and the first term (u1) is given as 36.

So, we can set up the equation:

u5 - u1 = 36

Since the fifth term is greater than the first term, we know that the common difference will be positive. Let's assume the common difference is represented as 'd'.

Therefore, u1 + 4d - u1 = 36 (since the fifth term is u1 + 4d)

Simplifying the equation, we get:

4d = 36

Dividing both sides by 4, we find:

d = 9

So, the common difference in this arithmetic sequence is 9.

b) To find u1, we are given that the tenth term of the sequence is double the seventh term. Let's assume the first term is represented as 'u1'. Based on this information, we can set up the equation:

u7 * 2 = u10

Since the seventh term is u1 + 6d (where d is the common difference), and the tenth term is u1 + 9d, we can rewrite the equation as:

(u1 + 6d) * 2 = (u1 + 9d)

Simplifying the equation, we get:

2u1 + 12d = u1 + 9d

Subtracting u1 from both sides:

u1 = -3d

Since we know the common difference is 9 from part a, we substitute d = 9 into the equation:

u1 = -3 * 9 = -27

Therefore, u1 is -27.

c) To calculate the value of 'a' in the given cubic function, we need to use the fact that the function has a local maximum at x = -2. At a local maximum (or minimum), the derivative of the function is equal to zero.

a) To find f'(x), we need to take the derivative of f(x) with respect to x. In this case, the derivative of ax^3 is 3ax^2, and the derivative of -3x is -3. The derivative of the constant term 5 is zero. So, combining these, we get:

f'(x) = 3ax^2 - 3

b) To find the value of f'(0), we substitute x = 0 into the derivative:

f'(0) = 3a(0)^2 - 3
= 0 - 3
= -3

Therefore, the value of f'(0) is -3.

c) At the local maximum (x = -2), the derivative f'(x) should equal zero. So, we substitute x = -2 into the derivative:

3a(-2)^2 - 3 = 0
12a - 3 = 0
12a = 3
a = 3/12
a = 0.25

Therefore, the value of a is 0.25.

#1

(a) ok
(b)
u10 = a+9d = a+81
u7 = a+6d = a+54
a+81 = 2(a+54)
a = -27

#2 ok