iGCSE Mathematics (0580) - C2.7 Sequences- Exam Style Questions Paper 1- New Syllabus
▶️ Answer/Explanation
(a)
The common difference is \(8 – 2 = 6\), so the sequence is linear with \(d = 6\). The \(n\)th term formula is \(a + (n-1)d = 2 + (n-1) \times 6 = 6n – 4\).
Checking: \(n=1 \Rightarrow 2\), \(n=2 \Rightarrow 8\), \(n=3 \Rightarrow 14\), \(n=4 \Rightarrow 20\). ✓
✅ Answer: \(6n – 4\)
(b)
Substitute \(n = 1, 2, 3\) into \(n^2 + 10\):
\(n=1:\ 1^2 + 10 = 11\); \(\quad n=2:\ 2^2 + 10 = 14\); \(\quad n=3:\ 3^2 + 10 = 19\).
✅ Answer: \(11,\quad 14,\quad 19\)
▶️ Answer/Explanation
(a)
Observe the difference between consecutive terms: \(8-1=7\), \(15-8=7\). Adding \(7\) to the last term \(22\) gives \(29\).
✅ Answer: \(29\)
(b)
As established above, each term increases by \(7\).
✅ Answer: Add 7
(c)(i)
Because the common difference is \(7\), the \(n\)th term has the form \(7n + c\).
For the first term (\(n=1\)), \(7(1) + c = 1 \implies c = -6\). Wait, the first term is \(1\), so \(7(1) – 6 = 1\).
✅ Answer: \(7n – 6\) (Correction from provided MS rule: \(7n-6\))
(c)(ii)
Set the \(n\)th term equal to \(688\): \(7n – 6 = 688\).
\(7n = 694\). Check if \(694\) is divisible by \(7\). \(694 \div 7 \approx 99.14\), which is not a whole number.
✅ Answer: No, because 694 is not perfectly divisible by 7.
▶️ Answer/Explanation
(a)(ii) Continuing this subtraction rule from the last term ($12$), the next term is $12 – 7 = 5$, and the one following that is $5 – 7 = -2$.
(b) Let’s check the step difference for this second sequence: $23 – 19 = +4$, $27 – 23 = +4$, so it goes up by $4$ each time. This tells us the formula starts with $4n$. To find the constant shift, check when $n=1$: $4(1) = 4$, but our first term is $19$. We need to add $15$ to get there ($4 + 15 = 19$). Thus, the full general expression is $4n + 15$.
✅ Answer:
(a)(i) Subtract $7$ (or $-7$)
(a)(ii) $5$ and $-2$
(b) $4n + 15$