Sequences and series
Sequences and series
- A2
- Non-calculator
- 8 questions
- 37 marks
These are original practice questions for the Edexcel A-level Pure content, not a past paper. This site is independent and not affiliated with Pearson.
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An arithmetic sequence has first term $5$ and common difference $3$. Find the $20$th term.
3 marks
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A geometric sequence has first term $4$ and common ratio $\dfrac{1}{2}$. Find the $6$th term.
3 marks
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Find the sum of the first $15$ terms of the arithmetic series $7 + 11 + 15 + \cdots$.
4 marks
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Find the sum of the first $8$ terms of the geometric series $3 + 6 + 12 + \cdots$.
4 marks
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Find the sum to infinity of the series $12 + 8 + \dfrac{16}{3} + \cdots$.
5 marks
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Evaluate $\displaystyle \sum_{r=1}^{20} (3r + 1)$.
5 marks
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An arithmetic sequence has $4$th term $15$ and $9$th term $35$. Find the first term and the sum of the first $20$ terms.
6 marks
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A geometric series has second term $6$ and sum to infinity $27$. Find the two possible values of the common ratio, and the corresponding first terms.
7 marks
Mark scheme
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Answer. $62$
- $u_{20} = 5 + 19 \times 3 = 62$.
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Answer. $\dfrac{1}{8}$
- $u_6 = 4 \times \left(\dfrac{1}{2}\right)^5 = \dfrac{4}{32} = \dfrac{1}{8}$.
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Answer. $525$
- The first term is $7$ and the common difference is $4$.
- $S_{15} = \dfrac{15}{2}\big(2 \times 7 + 14 \times 4\big) = 525$.
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Answer. $765$
- The first term is $3$ and the common ratio is $2$.
- $S_8 = 3(2^8 - 1) = 3 \times 255 = 765$.
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Answer. $36$
- The common ratio is $\dfrac{2}{3}$, and $\left|\dfrac{2}{3}\right| < 1$.
- $S_{\infty} = \dfrac{12}{1 - \dfrac{2}{3}} = 36$.
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Answer. $650$
- $\displaystyle \sum_{r=1}^{20} (3r + 1) = 3 \times \dfrac{20 \times 21}{2} + 20$.
- $630 + 20 = 650$.
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Answer. The first term is $3$ and the sum is $820$.
- $a + 3d = 15$ and $a + 8d = 35$.
- $5d = 20$, so $d = 4$ and $a = 3$.
- $S_{20} = 10\big(6 + 19 \times 4\big) = 820$.
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Answer. The common ratio is $\dfrac{1}{3}$ or $\dfrac{2}{3}$. The first terms are $18$ and $9$.
- $ar = 6$ and $\dfrac{a}{1 - r} = 27$, with $|r| < 1$.
- $\dfrac{6}{r(1 - r)} = 27$, so $9r - 9r^2 = 2$.
- $9r^2 - 9r + 2 = 0$, so $(3r - 1)(3r - 2) = 0$.
- $r = \dfrac{1}{3}$ gives $a = 18$, and $r = \dfrac{2}{3}$ gives $a = 9$.