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# SERIES AND SEQUENCE QUESTION – EXAM SOLUTION

## 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 ratioi) 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

YOU CAN DO IT

### Oluwamuyide Peter

On the 4th of November I officially became a member of the exclusive 1st student with distinction after five years of no such record, in the history of The Polytechnic Ibadan, Faculty of Engineering to graduate with distinction as a DPP students since its establishment in 2011. My unrelenting power to solve problems, have made me to create a platform where student can get valid information anywhere, anyplace at anytime Evolving education world wide 🌎