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Arithmetic & Geometric Series — Easy Worksheet 1
Grade 11 · Whoobee Name: ____________________ Date: __________
Find each sum. Write fractions in lowest terms.
How to solve
Arithmetic series: S_n = n(a_1 + a_n) ÷ 2, where a_n = a_1 + (n - 1)d. Geometric series: S_n = a_1(1 - rn) ÷ (1 - r). An infinite geometric series with |r| < 1 adds up to a_1 ÷ (1 - r). In sigma notation, substitute k = 1, 2, … up to the top number and add the terms.
Example
Find the sum of the first 10 terms of the arithmetic series 10 + 14 + 18 + ….
Answer: 280
Not quite. The answer is
💡 The first term is a_1 = −4 and the common difference is d = −1 − (−4) = 3. The last term is a_10 = a_1 + (n − 1)d = −4 + 9(3) = 23. Use S_n = n(a_1 + a_n) ÷ 2: S_10 = 10(−4 + 23) ÷ 2 = 10 × 19 ÷ 2 = 95. The sum is 95.
✓ Correct!
Not quite. The answer is
💡 The first term is a_1 = −4 and the common difference is d = −1 − (−4) = 3. The last term is a_10 = a_1 + (n − 1)d = −4 + 9(3) = 23. Use S_n = n(a_1 + a_n) ÷ 2: S_10 = 10(−4 + 23) ÷ 2 = 10 × 19 ÷ 2 = 95. The sum is 95.
💡 How to solve: Arithmetic & Geometric Series
Arithmetic series: S_n = n(a_1 + a_n) ÷ 2, where a_n = a_1 + (n - 1)d. Geometric series: S_n = a_1(1 - rn) ÷ (1 - r). An infinite geometric series with |r| < 1 adds up to a_1 ÷ (1 - r). In sigma notation, substitute k = 1, 2, … up to the top number and add the terms.
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Find the sum of the first 10 terms of the arithmetic series 0 + 3 + 6 + ….
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Not quite. The answer is
💡 Put in k = 1, 2, 3, …: the terms are 0, −2, −4, … , and k = 4 gives the last term −6. They go down by 2 each time, so this is an arithmetic series with 4 terms. Use S_n = n(a_1 + a_n) ÷ 2: S_4 = 4(0 + (−6)) ÷ 2 = 4 × (−6) ÷ 2 = −12. The sum is −12.
✓ Correct!
Not quite. The answer is
💡 Put in k = 1, 2, 3, …: the terms are 0, −2, −4, … , and k = 4 gives the last term −6. They go down by 2 each time, so this is an arithmetic series with 4 terms. Use S_n = n(a_1 + a_n) ÷ 2: S_4 = 4(0 + (−6)) ÷ 2 = 4 × (−6) ÷ 2 = −12. The sum is −12.
💡 How to solve: Arithmetic & Geometric Series
Arithmetic series: S_n = n(a_1 + a_n) ÷ 2, where a_n = a_1 + (n - 1)d. Geometric series: S_n = a_1(1 - rn) ÷ (1 - r). An infinite geometric series with |r| < 1 adds up to a_1 ÷ (1 - r). In sigma notation, substitute k = 1, 2, … up to the top number and add the terms.
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Find the sum of the first 11 terms of the arithmetic series -3 + 1 + 5 + ….
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Not quite. The answer is
💡 The first term is a_1 = 9 and the common difference is d = 11 − 9 = 2. The last term is a_7 = a_1 + (n − 1)d = 9 + 6(2) = 21. Use S_n = n(a_1 + a_n) ÷ 2: S_7 = 7(9 + 21) ÷ 2 = 7 × 30 ÷ 2 = 105. The sum is 105.
✓ Correct!
Not quite. The answer is
💡 The first term is a_1 = 9 and the common difference is d = 11 − 9 = 2. The last term is a_7 = a_1 + (n − 1)d = 9 + 6(2) = 21. Use S_n = n(a_1 + a_n) ÷ 2: S_7 = 7(9 + 21) ÷ 2 = 7 × 30 ÷ 2 = 105. The sum is 105.
💡 How to solve: Arithmetic & Geometric Series
Arithmetic series: S_n = n(a_1 + a_n) ÷ 2, where a_n = a_1 + (n - 1)d. Geometric series: S_n = a_1(1 - rn) ÷ (1 - r). An infinite geometric series with |r| < 1 adds up to a_1 ÷ (1 - r). In sigma notation, substitute k = 1, 2, … up to the top number and add the terms.
Try a similar one
Find the sum of the first 7 terms of the arithmetic series 16 + 20 + 24 + ….
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Not quite. The answer is
💡 The first term is a_1 = 5 and the common ratio is r = 10 ÷ 5 = 2. Use S_n = a_1(1 − rn) ÷ (1 − r): S_6 = 5(1 − 26) ÷ (1 − 2) = 5(1 − 64) ÷ (−1) = −315 ÷ (−1) = 315. The sum is 315.
✓ Correct!
Not quite. The answer is
💡 The first term is a_1 = 5 and the common ratio is r = 10 ÷ 5 = 2. Use S_n = a_1(1 − rn) ÷ (1 − r): S_6 = 5(1 − 26) ÷ (1 − 2) = 5(1 − 64) ÷ (−1) = −315 ÷ (−1) = 315. The sum is 315.
💡 How to solve: Arithmetic & Geometric Series
Arithmetic series: S_n = n(a_1 + a_n) ÷ 2, where a_n = a_1 + (n - 1)d. Geometric series: S_n = a_1(1 - rn) ÷ (1 - r). An infinite geometric series with |r| < 1 adds up to a_1 ÷ (1 - r). In sigma notation, substitute k = 1, 2, … up to the top number and add the terms.
Try a similar one
Find the sum of the first 9 terms of the arithmetic series 0 + 4 + 8 + ….
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Not quite. The answer is
💡 The first term is a_1 = 1 and the common difference is d = 6 − 1 = 5. The last term is a_9 = a_1 + (n − 1)d = 1 + 8(5) = 41. Use S_n = n(a_1 + a_n) ÷ 2: S_9 = 9(1 + 41) ÷ 2 = 9 × 42 ÷ 2 = 189. The sum is 189.
✓ Correct!
Not quite. The answer is
💡 The first term is a_1 = 1 and the common difference is d = 6 − 1 = 5. The last term is a_9 = a_1 + (n − 1)d = 1 + 8(5) = 41. Use S_n = n(a_1 + a_n) ÷ 2: S_9 = 9(1 + 41) ÷ 2 = 9 × 42 ÷ 2 = 189. The sum is 189.
💡 How to solve: Arithmetic & Geometric Series
Arithmetic series: S_n = n(a_1 + a_n) ÷ 2, where a_n = a_1 + (n - 1)d. Geometric series: S_n = a_1(1 - rn) ÷ (1 - r). An infinite geometric series with |r| < 1 adds up to a_1 ÷ (1 - r). In sigma notation, substitute k = 1, 2, … up to the top number and add the terms.
Try a similar one
Find the sum of the first 6 terms of the arithmetic series 5 + 7 + 9 + ….
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 Put in k = 1, 2, 3, …: the terms are 3, 6, 9, … , and k = 4 gives the last term 12. They go up by 3 each time, so this is an arithmetic series with 4 terms. Use S_n = n(a_1 + a_n) ÷ 2: S_4 = 4(3 + 12) ÷ 2 = 4 × 15 ÷ 2 = 30. The sum is 30.
✓ Correct!
Not quite. The answer is
💡 Put in k = 1, 2, 3, …: the terms are 3, 6, 9, … , and k = 4 gives the last term 12. They go up by 3 each time, so this is an arithmetic series with 4 terms. Use S_n = n(a_1 + a_n) ÷ 2: S_4 = 4(3 + 12) ÷ 2 = 4 × 15 ÷ 2 = 30. The sum is 30.
💡 How to solve: Arithmetic & Geometric Series
Arithmetic series: S_n = n(a_1 + a_n) ÷ 2, where a_n = a_1 + (n - 1)d. Geometric series: S_n = a_1(1 - rn) ÷ (1 - r). An infinite geometric series with |r| < 1 adds up to a_1 ÷ (1 - r). In sigma notation, substitute k = 1, 2, … up to the top number and add the terms.
Try a similar one
Find the sum of the first 9 terms of the arithmetic series 3 + 5 + 7 + ….
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 The first term is a_1 = 10 and the common difference is d = 12 − 10 = 2. The last term is a_11 = a_1 + (n − 1)d = 10 + 10(2) = 30. Use S_n = n(a_1 + a_n) ÷ 2: S_11 = 11(10 + 30) ÷ 2 = 11 × 40 ÷ 2 = 220. The sum is 220.
✓ Correct!
Not quite. The answer is
💡 The first term is a_1 = 10 and the common difference is d = 12 − 10 = 2. The last term is a_11 = a_1 + (n − 1)d = 10 + 10(2) = 30. Use S_n = n(a_1 + a_n) ÷ 2: S_11 = 11(10 + 30) ÷ 2 = 11 × 40 ÷ 2 = 220. The sum is 220.
💡 How to solve: Arithmetic & Geometric Series
Arithmetic series: S_n = n(a_1 + a_n) ÷ 2, where a_n = a_1 + (n - 1)d. Geometric series: S_n = a_1(1 - rn) ÷ (1 - r). An infinite geometric series with |r| < 1 adds up to a_1 ÷ (1 - r). In sigma notation, substitute k = 1, 2, … up to the top number and add the terms.
Try a similar one
Find the sum of the first 10 terms of the arithmetic series 6 + 10 + 14 + ….
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 The first term is a_1 = 1 and the common ratio is r = 3 ÷ 1 = 3. Use S_n = a_1(1 − rn) ÷ (1 − r): S_5 = (1 − 35) ÷ (1 − 3) = (1 − 243) ÷ (−2) = −242 ÷ (−2) = 121. The sum is 121.
✓ Correct!
Not quite. The answer is
💡 The first term is a_1 = 1 and the common ratio is r = 3 ÷ 1 = 3. Use S_n = a_1(1 − rn) ÷ (1 − r): S_5 = (1 − 35) ÷ (1 − 3) = (1 − 243) ÷ (−2) = −242 ÷ (−2) = 121. The sum is 121.
💡 How to solve: Arithmetic & Geometric Series
Arithmetic series: S_n = n(a_1 + a_n) ÷ 2, where a_n = a_1 + (n - 1)d. Geometric series: S_n = a_1(1 - rn) ÷ (1 - r). An infinite geometric series with |r| < 1 adds up to a_1 ÷ (1 - r). In sigma notation, substitute k = 1, 2, … up to the top number and add the terms.
Try a similar one
Find the sum of the first 12 terms of the arithmetic series 4 + 7 + 10 + ….
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
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Arithmetic & Geometric Series — Easy Worksheet 1
Find each sum. Write fractions in lowest terms.