By Fatskills Exam Guides Team — the exam nerds behind 28,500+ quizzes and 2.1M practice questions across 500+ global exams.
GCSE & A-Level Maths
"Mastering sequences and series doesn’t just get you 10–15 marks on your exam—it’s the key to predicting loan repayments, calculating investment growth, and even cracking coding algorithms. One wrong formula, and your answer could be off by thousands. Let’s make sure you nail it."
MEMORISE THIS
Sum of first n terms: ( S_n = \frac{n}{2} (2a_1 + (n-1)d) ) OR ( S_n = \frac{n}{2} (a_1 + a_n) )
Sum of first n terms: ( S_n = \frac{a_1 (1 - r^n)}{1 - r} ) (if ( r \neq 1 ))
Sum to infinity: ( S_\infty = \frac{a_1}{1 - r} ) (only if ( |r| < 1 ))
Geometric: Ratio between terms is constant.
Write down what you know – First term (( a_1 )), common difference (( d )) or ratio (( r )), number of terms (( n )).
Pick the right formula – nth term, sum, or sum to infinity?
Substitute and solve – Plug numbers into the formula.
Check your answer – Does it make sense? (e.g., sum can’t be negative if all terms are positive).
Question: An arithmetic sequence has first term 5 and common difference 3. Find the 10th term and the sum of the first 10 terms.
Step 1: It’s arithmetic (difference = 3). Step 2: ( a_1 = 5 ), ( d = 3 ), ( n = 10 ). Step 3: - For the 10th term: ( a_n = a_1 + (n-1)d ) - For the sum: ( S_n = \frac{n}{2} (2a_1 + (n-1)d) ) Step 4: - 10th term: ( a_{10} = 5 + (10-1) \times 3 = 5 + 27 = 32 ) - Sum: ( S_{10} = \frac{10}{2} (2 \times 5 + 9 \times 3) = 5 (10 + 27) = 5 \times 37 = 185 ) Step 5: Check – 10th term is reasonable (starts at 5, adds 3 each time). Sum is positive, as expected.
Question: A sequence starts 7, 11, 15, 19… Find the 20th term and the sum of the first 20 terms.
Solution: 1. Identify: Arithmetic (difference = 4). 2. Known: ( a_1 = 7 ), ( d = 4 ), ( n = 20 ). 3. Formulas: - nth term: ( a_n = a_1 + (n-1)d ) - Sum: ( S_n = \frac{n}{2} (2a_1 + (n-1)d) ) 4. Substitute: - 20th term: ( a_{20} = 7 + 19 \times 4 = 7 + 76 = 83 ) - Sum: ( S_{20} = \frac{20}{2} (14 + 19 \times 4) = 10 (14 + 76) = 10 \times 90 = 900 ) 5. Check: 20th term is ( 7 + 19 \times 4 = 83 ) (correct). Sum is positive.
What we did and why: We used the arithmetic formulas because the sequence adds 4 each time. Always write down the known values first.
Question: A geometric sequence has first term 3 and common ratio 2. Find the 6th term and the sum of the first 6 terms.
Solution: 1. Identify: Geometric (ratio = 2). 2. Known: ( a_1 = 3 ), ( r = 2 ), ( n = 6 ). 3. Formulas: - nth term: ( a_n = a_1 \times r^{n-1} ) - Sum: ( S_n = \frac{a_1 (1 - r^n)}{1 - r} ) 4. Substitute: - 6th term: ( a_6 = 3 \times 2^{5} = 3 \times 32 = 96 ) - Sum: ( S_6 = \frac{3 (1 - 2^6)}{1 - 2} = \frac{3 (1 - 64)}{-1} = \frac{3 \times -63}{-1} = 189 ) 5. Check: 6th term is ( 3 \times 32 = 96 ) (correct). Sum is positive.
What we did and why: We used geometric formulas because the sequence multiplies by 2 each time. Remember: ( r^n ) means ( r ) to the power of ( n ), not ( r \times n ).
Question: Evaluate ( \sum_{r=1}^{5} (2r + 1) ).
Solution: 1. Identify: Sigma notation means sum the expression ( 2r + 1 ) from ( r = 1 ) to ( r = 5 ). 2. Write out the terms: - ( r = 1 ): ( 2(1) + 1 = 3 ) - ( r = 2 ): ( 2(2) + 1 = 5 ) - ( r = 3 ): ( 2(3) + 1 = 7 ) - ( r = 4 ): ( 2(4) + 1 = 9 ) - ( r = 5 ): ( 2(5) + 1 = 11 ) 3. Sum them: ( 3 + 5 + 7 + 9 + 11 = 35 ) 4. Alternative (using arithmetic sum): - First term (( a_1 )) = 3, last term (( a_5 )) = 11, ( n = 5 ). - Sum = ( \frac{n}{2} (a_1 + a_n) = \frac{5}{2} (3 + 11) = \frac{5}{2} \times 14 = 35 ) 5. Check: Both methods give 35.
What we did and why: Sigma notation is just a shorthand for adding terms. Writing them out avoids mistakes. The arithmetic sum formula is faster for large ( n ).
CORRECT APPROACH: Always check if the sequence adds (arithmetic) or multiplies (geometric).
Mistake: Forgetting ( n-1 ) in the nth term formula.
CORRECT APPROACH: Write the formula down every time.
Mistake: Using the sum to infinity formula when ( |r| \geq 1 ).
CORRECT APPROACH: Only use ( S_\infty = \frac{a_1}{1 - r} ) if ( |r| < 1 ).
Mistake: Misinterpreting sigma notation (e.g., ( \sum_{r=1}^{3} r^2 ) as ( (1+2+3)^2 )).
CORRECT APPROACH: Write out the terms: ( 1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14 ).
Mistake: Sign errors in the geometric sum formula.
How to Avoid it: Check the difference/ratio between terms before choosing a formula.
Trap: Sum to infinity questions where ( |r| \geq 1 ).
How to Avoid it: Remember: sum to infinity only exists if ( |r| < 1 ). If not, write "diverges" or "no sum."
Trap: Sigma notation with a lower limit not equal to 1.
"Here’s the night-before cheat sheet: 1. Arithmetic sequences: Add the same number each time. Use ( a_n = a_1 + (n-1)d ) for the nth term and ( S_n = \frac{n}{2} (2a_1 + (n-1)d) ) for the sum. 2. Geometric sequences: Multiply by the same number each time. Use ( a_n = a_1 \times r^{n-1} ) and ( S_n = \frac{a_1 (1 - r^n)}{1 - r} ). For sum to infinity, ( S_\infty = \frac{a_1}{1 - r} ) only if ( |r| < 1 ). 3. Sigma notation: Just a fancy way to write sums. Write out the terms if you’re unsure. 4. Common mistakes: Forgetting ( n-1 ), mixing up arithmetic/geometric, and sign errors in the geometric sum formula. 5. Exam traps: Check if the sequence is arithmetic or geometric, watch for ( |r| \geq 1 ) in sum to infinity, and don’t assume sigma notation starts at 1.
Memorise the formulas, practice one of each type tonight, and you’ll smash it tomorrow. Good luck!
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