Home / iGCSE Mathematics (0580) :E2.7 Continue a given number sequence.iGCSE Style Questions Paper 4

iGCSE Mathematics (0580) :E2.7 Continue a given number sequence.iGCSE Style Questions Paper 4

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IGCSE Mathematics (0580) Practice Questions Paper 4 - All Chapters
Question

(a) The nth term of a sequence is \( 120 – n^3 \).

(a) (i) Find the 4th term of this sequence.

(a) (ii) Find the value of \( n \) when the nth term is \(-1211\).

(b) The nth term of a different sequence is \( 3 \times (0.2)^{n-1} \). Find the 5th term of this sequence.

(c) The table shows the first four terms of sequences \( A \), \( B \), and \( C \).

Sequence1st term2nd term3rd term4th term5th termnth term
\( A \)741-2  
\( B \)\(\frac{1}{4}\)\(\frac{2}{5}\)\(\frac{3}{6}\)\(\frac{4}{7}\)  
\( C \)02612  

Complete the table for each sequence.

▶️ Answer/Explanation
Solution

(a) (i) Ans: 56

The nth term is \( 120 – n^3 \). For the 4th term, substitute \( n = 4 \):

\( 120 – 4^3 = 120 – 64 = 56 \).

(a) (ii) Ans: 11

Set \( 120 – n^3 = -1211 \), solve for \( n \):

\( n^3 = 1331 \implies n = \sqrt[3]{1331} = 11 \).

(b) Ans: 0.0048 or \( \frac{3}{625} \)

The nth term is \( 3 \times (0.2)^{n-1} \). For the 5th term, substitute \( n = 5 \):

\( 3 \times (0.2)^4 = 3 \times 0.0016 = 0.0048 \).

(c)  

Sequence A: Decreases by 3 each time. 5th term = \(-5\), nth term = \( 10 – 3n \).

Sequence B: Numerator and denominator increase by 1. 5th term = \( \frac{5}{8} \), nth term = \( \frac{n}{n+3} \).

Sequence C: Follows \( n^2 – n \). 5th term = 20, nth term = \( n^2 – n \).

Question

These are the first four diagrams of a sequence. The diagrams are made from white dots and black dots.

Dot sequence diagram

(a) Complete the table for Diagram 5 and Diagram 6.

Dot sequence table

(b) Write an expression, in terms of n, for the number of white dots in Diagram n.

(c) The expression for the total number of dots in Diagram n is \(\frac{1}{2}(3n^{2}-n)\).

(i) Find the total number of dots in Diagram 8.

(ii) Find an expression for the number of black dots in Diagram n.

(d) T is the total number of dots used to make all of the first n diagrams. \(T=an^{3}+bn^{2}\) Find the value of a and b.

▶️ Answer/Explanation
Answer:
(a) 25, 36, 10, 15, 35, 51
(b) \(n^{2}\)
(c)(i) 92
(ii) \(\frac{1}{2}(n^{2}-n)\)
(d) \(a=\frac{1}{2}\), \(b=\frac{1}{2}\)
Detailed Solution:
(a) White dots follow squares (1,4,9,16 → 25,36). Black dots follow triangular numbers (1,3,6,10 → 15,21). Total = white + black.
(b) White dots = n² (since 1²=1, 2²=4, 3²=9, etc.).
(c)(i) Plug n=8 into \(\frac{1}{2}(3(8)^2 – 8) = \frac{1}{2}(192-8) = 92\).
(c)(ii) Black dots = Total dots – White dots = \(\frac{1}{2}(3n^2 – n) – n^2 = \frac{1}{2}(n^2 – n)\).
(d) Sum T = Σ Total dots from n=1 to n. Expand and simplify to match \(T = \frac{1}{2}n^3 + \frac{1}{2}n^2\).
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