First year = $1,560

Second year = $1,560 + y

Third year = $1,560 + y + y

Fourth year = $1,560 + y + y + y

This is a linear sequence: $1,560,$1,560 + y, $1,560 + 2y $1,560 + 3y,...,$13,980

The value of the nth term of a linear sequence is given by U₁ + (n-1)d

Where U₁ = the first term
n = the nth term
d = the common difference

In the above story problem, the first term is $1,560

The common difference = y

We were given the 13th term which is $13,980

We can therefore express the 13th term using the above formula and solve for the value of y

$13980 = $1560 + (13-1) x y

$13980 - $1560 = 12y

$12420 = 12y

Divide both sides by 12

y = $\frac{\mathrm{\$12420}}{12}$ = $1035

The sequence is linear.

Common difference = $\frac{5}{8}$ - $\frac{1}{4}$ = 1 - $\frac{5}{8}$ = 1$\frac{3}{8}$ - 1 = $\frac{3}{8}$

U_n = a + (n - 1)d

Where n = n^th term,
a = first term
d = common difference

a = $\frac{1}{4}$

d = $\frac{3}{8}$

U₁₁ = $\frac{1}{4}$ + (11 - 1) x $\frac{3}{8}$

U₁₁ = $\frac{1}{4}$ + 10 x $\frac{3}{8}$

U₁₁ = $\frac{1}{4}$ + 5 x $\frac{3}{4}$

U₁₁ = $\frac{1}{4}$ + $\frac{15}{4}$

U₁₁ = $\frac{\mathrm{1\; +\; 15}}{4}$

U₁₁ = $\frac{16}{4}$

U₁₁ = 4

For Arithmetic Progression (A.P), the common difference is the same.

Common difference = Succeeding - Preceding

6, p, 14, ...

14 - p = p - 6

Solve for p

14 + 6 = p + p

20 = 2p

Divide both sides by 2.

p = $\frac{20}{2}$ = 10

a₁ = -20

n = 4

a₄ = 37

-20 + (4 - 1) x d = 37

-20 + 3d = 37

3d = 37 + 20

3d = 57

Divide both sides by 3

d = $\frac{57}{3}$ = 19