**PROBLEM 1**: Given ∫ᵅ₋ₐ 15x²dx = 3430. Find the value of the constant a

∫ᵅ₋ₐ 15x²dx = 3430

dx = [5x³/3]ᵅ₋ₐ

[5(a)³/3] – [5(-a)³/3] = 3430

5a³ + 5a³ = 3430

10a³ = 3430

10a³/10 = 3430/10

a³ = 343

a = 3√343

a = 7

**PROBLEM 2: **Using the quadratic formular, determine the roots of the equation x²+6x+m = 0, ∀m ∈ R

what is the meaning of this ∀m ∈ R

∀ means for ALL

∈ means is a MEMBER

R means REAL NUMBER

∀m ∈ R means for ALL m are MEMBERS of a REAL NUMBER

LET’S SOLVE

x = [-b± √(b²-4ac) ] ÷ 2a

x²+6x+m = 0

Let A = 1, B = 6 and C = 0

x = [-6± √(6²-4×1×m) ] ÷ 2×1

x = [-6÷2 ± √(36-4m)÷2

x = [-3 ± √(-4[9-m]m) ÷ 2

x = -3 ± √(9-m); Therefore x = -3 + √(9-m) or x = -3 – √(9-m)

### Given that these roots differ by 2k, k∈R and K ≠ 0, show that k² = 9-m

Continuation of question 2

-3 + √(9-m) – ( -3 – √(9-m) = 2k

-3 + √(9-m) + 3 + √(9-m) = 2k

2√(9-m) = 2k

√(9-m) = k

K² = 9-m (PROVED)

**PROBLEM 3**: For a geometric progression (Vₙ), the sum of the first n terms is sₙ = V₁(1-rⁿ)÷(1-r); r ≠ 1 where r is the common ratio

i) Show that (S₃ₙ – S₂ₙ) ÷ Sₙ = r²ⁿ

Vn = Last term of Geometry progression

sₙ = V₁(1-rⁿ)÷(1-r); r ≠ 1

Note: V1 is the first term

S3n = V₁(1-r3n)÷(1-r), S2n = V₁(1-r2n)÷(1-r); and sₙ = V₁(1-rⁿ)÷(1-r)

Substitute it into (S₃ₙ – S₂ₙ) ÷ Sₙ

[ V₁(1-r3n)÷(1-r) – V₁(1-r2n)÷(1-r) ] ÷ [V₁(1-rⁿ)÷(1-r)][ V₁(1-r3n)÷(1-r) – V₁(1-r2n)÷(1-r) ] × [ (1-r)÷V₁(1-rⁿ) ](V1r2n – V1r3n) ÷ V1 (1-r n)

(V1r2n – V1r3n) ÷ ( V1 – V1 r n)

r2n (V1 – V1r3n) ÷ ( V1 – V1 r n)

⇒ r²ⁿ

### Hence find, for when r = ½

r²ⁿ, where n = 0, r = ½

r²ⁿ ⇒½²⁽⁰⁾ =½⁰ (Note anything raise to power zero is one)

= 1

**PROBLEM 4:** An arithmetic progression (Un) is defined ∀n ∈ ℕ

**Question 1**: Determine the first term U0 and the common difference of (Un)

U0 = a, U1= a+d, U2 = a+2d and U5= a+5d

a = first term, d = common difference

U0 + U1 + U2 = 15

and U5 = 17

a + ( a+d) + (a+2d) = 15

3a + 3d =15

a + d = 5 ….. equation (1)

and a+5d = 17…….equ equation (2)

Using elimination method

a + d = 5 …..(1)

-[ a + 5d = 17] …(2)

Common difference, d = 3

Substitute d as 3 into equation 1

a + 3 = 5

a = 5-3; = 2

**Question 2:** What is the sum of the first 25 terms

When a = 2 and d = 3

Sn = n/2 [ 2a+(n-1)d]S25 = 25/2 [ 2×2+(25-1)3 ]= 12.5 [ 4+(24×3) ]= 12.5 [ 4+72 ]= 12.5 × 76

⇒950