(i)

U_n = a + (n - 1) x d

U_n = the nth term

a = first term

n = position of term

d = common difference

fifth term = a + (5 - 1) x d

fifth term = a + 4 x d

fifth term = a + 4d

eighth term = a + (8 - 1) x d

eighth term = a + 7 x d

eighth term = a + 7d

fifth term is 6 less than the eighth term means if you add 6 to the fifth term, you should get the eighth term.

a + 4d + 6 = a + 7d

6 = a - a + 7d - 4d

6 = 3d

Divide both sides by 3

d = $\frac{6}{3}$ = 2

The common difference = 2

(ii)

S_n = $\frac{\mathrm{n}}{2}$[2a + (n - 1)d

n = 10

S_n = 120

$\frac{10}{2}$[2a + (10 - 1)d] = 120

5(2a + 9d) = 120

10a + 45d = 120

But d = 2

10a + 45 x 2 = 120

10a + 90 = 120

10a = 120 - 90

10a = 30

Divide both sides by 10

a = $\frac{30}{10}$ = 3

First term = 3

(a)

The common difference is the same. The common difference is succeeding - preceding

(6x – 3) - (4x – 2) = (4x – 2) - (x + 1)

6x – 3 - 4x + 2 = 4x – 2 - x - 1

2x - 1 = 3x - 3

3 - 1 = 3x - 2x

2 = x

x = 2

(b)

First term = (x + 1)
Second term = (4x – 2)
Third term = (6x – 3)

But x = 2

First term = (x + 1) = 2 + 1 = 3
Second term = (4x – 2) = 4 x 2 - 2 = 6
Third term = (6x – 3) = 6 x 2 - 3 = 9

a = 3

d = 6 - 3 = 9 - 6 = 3

Since we are given the last term to be 18, we can use it to find n (number of terms)

a_n = 18

a₁ = 3

18 = 3 + (n - 1) x 3

18 - 3 = 3n - 3

15 = 3n - 3

15 + 3 = 3n

18 = 3n

Divide both sides by 3

n = $\frac{18}{3}$ = 6

S_n = $\frac{\mathrm{n}}{2}[a\; +\; L]$

n = 6, a = 3 and L = 18

S_n = $\frac{6}{2}[3\; +\; 18]$

S_n = 3 x 21 = 63