MAT1501 Assignment 3 2025 - Due August 2025

MAT1501
ASSIGNMENT 3 2025
DUE: AUGUST 2025 (MEMO)

, MAT1501 Assessment 3 (2025)

Question 1
(a) A = {x ∈ R : x > −3} is all real numbers greater than −3. On a number line, draw an
open circle at −3 and shade to the right (toward +∞). B = (−4, 5] includes all x with
−4 < x ≤ 5. On a number line, draw an open circle at −4, a closed circle at 5, and shade
between them.

(b) By definition, A ∪ B = { x ∣ x ∈ A or x ∈ B} . Since B already includes all x >
−4 up to 5, and A includes all x > −3 beyond, their union is all x > −4. In set-builder
form:


A ∪ B = { x ∈ R ∣ x > −4}.

(c) The union A ∪ B is the interval (−4, ∞). Sketch this by an open circle at −4 and
shading to the right without bound.

(d) By definition, A ∩ B = {x ∣ x ∈ A and x ∈ B} . Here A requires x > −3 and B
requires −4 < x ≤ 5. Both conditions hold exactly when −3 < x ≤ 5. Thus


A ∩ B = { x ∈ R ∣ x > −3 and x ≤ 5},

or equivalently {x ∣ −3 < x ≤ 5}.

(e) The intersection A ∩ B is the interval (−3, 5]. On a number line, draw an open circle
at −3, a closed circle at 5, and shade between them.




Question 2
(a) Let the three arithmetic terms be a, a + d, a + 2d. We are given
a + (a + 2d) = 10, a(a + d) = 15.
From the sum: 2a + 2d = 10 ⟹ a + d = 5. From the product: a(a + d) = 15.
Substitute a + d = 5: a ⋅ 5 = 15 ⟹ a = 3. Then d = 5 − a = 2. The three terms are
3, 5, 7.

This satisfies 3 + 7 = 10 and 3 ⋅ 5 = 15. (An arithmetic sequence with first term 3 and
common difference 2 is 3, 5, 7.)


(b) We want Sn ​ = x2 − x3 + x4 − x5 + ⋯ (first n terms). Observe this is a geometric series:
factor x2 ,

n−2
2 2 3 n−2
Sn = x (1 − x + x − x + ⋯ + (−x)
​
) = x ∑ (−x)k .
2
​




k=0