Grade 12 Maths · Number Patterns, Sequences and Series · Find how many terms give a sum

Find how many terms give a sum Grade 12

Find n when a series adds up to a given total by setting S_n equal to it and solving (a quadratic for arithmetic, logs for geometric), or find the smallest n for which the sum passes a value.

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How many terms of the arithmetic sequence 5, 7, 9, …, must be added to get 140?AS Tutorial

Episode 12: How Many Terms of an Arithmetic Sequence are Required to Equal a Given Sum?PreMath

Geometric sequence(How to use logarithms) grade 12ALAN THE MATHS TUTOR

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Question 2.2 · May/June 2024 · Paper 1 · National (DBE) · 5 marks

2.2 Given:

\sum_{p = k}^{10} 3^{p - 1} = 29520

. Calculate the value of

k

.(5)

Work it out on paper first. Then mark yourself with the official memo, one tick at a time.

More past paper questions: find how many terms give a sum

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3. Question 3.1.2 · November 2023 · Paper 1 · National (DBE) · 4 marks
3.1 Given the geometric series: $3 + 6 + 12 + \dots$ to $n$ terms.

Answer the questions that follow.
3.1.2 Calculate the value of $k$ such that: $\sum_{p = 1}^{k} \frac{3}{2} (2)^{p} = 98301$ (4)
Show the memo (4 ticks)Memo
1. ✓
  Expansion $\sum_{p = 1}^{k} \frac{3}{2} (2)^{p} = 3 + 6 + 12 + \dots$
  1 mark · Read and write sigma notation
2. ✓
  $n = k$ $n = k$
  1 mark · Read and write sigma notation
3. ✓
  Substitution into correct formula $\frac{3 [( 2 ) ^{k} - 1 ]}{2 - 1} = 98301$
  1 mark · Find how many terms give a sum
4. ✓
  $k = 15$ $(2)^{k} - 1 = 32767$
  $(2)^{k} = 32768$
  $2^{k} = 2^{15}$ OR $k = lo g_{2} 32768$
  $∴ k = 15$
  1 mark · Find how many terms give a sum
4. Question 2.2.2 · November 2025 · Paper 1 · National (DBE) · 5 marks
2.2 Given $\sum_{p = k}^{117} (4 p - 1) = 26675$

Answer the questions that follow.
2.2.2 Calculate the value of $k$ .(5)
Show the memo (5 ticks)Memo
1. ✓
  METHOD 1: Number of terms $n = 118 - k$
  1 mark · Read and write sigma notation
2. ✓
  METHOD 1: Last term $T_{117} = 4 (117) - 1 = 467$
  1 mark · Read and write sigma notation
3. ✓
  METHOD 1: Substitution $S_{118 - k} = \frac{118 - k}{2} [4 k - 1 + 467]$
  $26675 = (118 - k) [2 k + 233]$
  1 mark · Find how many terms give a sum
4. ✓
  METHOD 1: Standard form $26675 = 236 k + 27494 - 2 k^{2} - 233 k$
  $2 k^{2} - 3 k - 819 = 0$
  1 mark · Write a quadratic in standard form
5. ✓
  METHOD 1: Answer with selection $(k - 21) (2 k + 39) = 0$
  $∴ k = 21$ or $k \neq = - \frac{39}{2}$
  1 mark · Find how many terms give a sum
5. Question 2.1.2 · November 2022 · Paper 1 · National (DBE) · 4 marks
2.1 The first term of a geometric series is $14$ and the $6^{th}$ term is $448$ .

Answer the following.
2.1.2 Determine the number of consecutive terms that must be added to the first $6$ terms of the series in order to obtain a sum of $114674$ .(4)
Show the memo (4 ticks)Memo
1. ✓
  METHOD 1: Substitution into the correct formula $T_{n} = 14 (2)^{n - 1}$
  $S_{n} = \frac{14 ( 2 ^{6} - 1 )}{2 - 1}$
  1 mark · Calculate a geometric series sum
2. ✓
  METHOD 1: $S_{6} = 882$ $S_{6} = 882$
  1 mark · Calculate a geometric series sum
3. ✓
  METHOD 1: $128 = 2^{n}$ $114674 - 882 = 113792$
  $113792 = 896 (2^{n} - 1)$
  $128 = 2^{n}$
  1 mark · Find how many terms give a sum
4. ✓
  METHOD 1: $7$ $n = 7$
  1 mark · Find how many terms give a sum
6. Question 2.1.4 · September 2026 · Paper 1 · Provincial · 4 marks
2.1 Given the arithmetic series: $7 + 16 + 25 + 34 + ...$
2.1.4 Determine the maximum number of terms for which the sum of this series will be less than $8600$ .(4)
Show the memo (4 ticks)Memo
1. ✓
  Equating $8600$ to $S_{n}$ $8600 = \frac{n}{2} [2 (7) + (n - 1) (9)]$
  1 mark · Find how many terms give a sum
2. ✓
  Simplification $17200 = n (14 + 9 n - 9)$
  $17200 = 9 n^{2} + 5 n$
  $9 n^{2} + 5 n - 17200 = 0$
  1 mark · Write a quadratic in standard form
3. ✓
  Substitution $n = \frac{- 5 \pm ( 5 ) ^{2} - 4 ( 9 ) ( - 17 200 )}{2 ( 9 )}$
  1 mark · Use the quadratic formula
4. ✓
  Answer $n = 43, 44$ or $n = - 43, 99$
  $∴ n = 43$
  1 mark · Find how many terms give a sum

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