Progression | Jamb Mathematics

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Cadet, listen up! Your next mission is to master the art of progression—arithmetic, geometric, and beyond—because precision and strategy are key to victory. Drill through formulas, decode patterns, and sharpen your problem-solving skills, because only the well-prepared survive the battlefield of mathematics. Fall in and get to work! 🚀🔥

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Are you preparing for your JAMB Mathematics exam and feeling a bit uncertain about how to approach the topic of Progression? Don’t worry—you’re in the right place! This lesson is here to break it down in a simple, clear, and engaging way, helping you build the strong foundation you need to succeed. Whether you're struggling with complex questions or just seeking a quick refresher, this guide will boost your understanding and confidence. Let’s tackle Progression together and move one step closer to achieving your exam success! Blissful learning.

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Calculation problem involving arithmetic progression

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1. Find the 10th term of the AP: 2, 5, 8, 11, ...

Solution: The formula for the nth term of an AP is:

a_n = a + (n-1) d

Where:

$a = 2$ (first term),
$d = 5 - 2 = 3$ (common difference),
$n = 10$ .

a_{10} = 2 + (10 - 1) \cdot 3

= 2 + 9 \cdot 3

= 2 + 27 = 29

Answer:

29

2. Find the sum of the first 15 terms of the AP: 3, 7, 11, 15, ...

Solution: Sum of the first

n

terms of an AP is given by:

S_n = \frac{n}{2} (2a + (n-1)d)

Here,

$a = 3$ ,
$d = 7 - 3 = 4$ ,
$n = 15$ .

S_{15} = \frac{15}{2} (2 \cdot 3 + (15 - 1) \cdot 4)

= \frac{15}{2} (6 + 56)

= \frac{15}{2} \cdot 62

= 15 \cdot 31

= 465

Answer:

465

3. The 7th term of an AP is 20, and the 13th term is 50. Find the first term and common difference.

Solution: Using the formula:

a_n = a + (n-1)d

For

a_7 = 20

a + 6d = 20

For

a_{13} = 50

a + 12d = 50

Subtracting equations:

(a + 12d) - (a + 6d) = 50 - 20

6d = 30

d = 5

Substituting

d

a + 6(5) = 20

a + 30 = 20

a = -10

Answer:

a = -10

d = 5

4. Find the number of terms in the AP: 4, 10, 16, ..., 88.

Solution: Using

a_n = a + (n-1)d

$a = 4$ ,
$d = 10 - 4 = 6$ ,
$a_n = 88$ .

88 = 4 + (n-1)6

88 - 4 = (n-1)6

84 = (n-1)6

n - 1 = 14

n = 15

Answer:

15

5. The sum of the first 20 terms of an AP is 1000. The first term is 7. Find the common difference.

Solution: Using

S_n = \frac{n}{2} (2a + (n-1)d)

1000 = \frac{20}{2} (2 \cdot 7 + (20-1)d)

1000 = 10 (14 + 19d)

100 = 14 + 19d

86 = 19d

d = \frac{86}{19} = 4.526

Answer:

d = 4.526

6. Find the 12th term of the AP: 6, 13, 20, 27, ...

Solution: Using

a_n = a + (n-1)d

$a = 6$ ,
$d = 13 - 6 = 7$ ,
$n = 12$ .

a_{12} = 6 + (12-1) \cdot 7

= 6 + 11 \cdot 7

= 6 + 77 = 83

Answer:

83

7. Find the sum of the first 25 terms of the AP: 2, 8, 14, 20, ...

Solution: Using

S_n = \frac{n}{2} (2a + (n-1)d)

$a = 2$ ,
$d = 8 - 2 = 6$ ,
$n = 25$ .

S_{25} = \frac{25}{2} (2 \cdot 2 + (25-1) \cdot 6)

= \frac{25}{2} (4 + 144)

= \frac{25}{2} \cdot 148

= 25 \cdot 74

= 1850

Answer:

1850

8. Find the number of terms in the AP: 3, 7, 11, ..., 87.

Solution: Using

a_n = a + (n-1)d

$a = 3$ ,
$d = 7 - 3 = 4$ ,
$a_n = 87$ .

87 = 3 + (n-1)4

87 - 3 = (n-1)4

84 = (n-1)4

n - 1 = 21

n = 22

Answer:

22

9. The 8th term of an AP is 36, and the 16th term is 84. Find the first term and common difference.

Solution: Using

a_n = a + (n-1)d

For

a_8 = 36

a + 7d = 36

For

a_{16} = 84

a + 15d = 84

Subtracting equations:

(a + 15d) - (a + 7d) = 84 - 36

8d = 48

d = 6

Substituting

d

a + 7(6) = 36

a + 42 = 36

a = -6

Answer:

a = -6

d = 6

10. The sum of the first 30 terms of an AP is 2400. The first term is 10. Find the common difference.

Solution: Using

S_n = \frac{n}{2} (2a + (n-1)d)

2400 = \frac{30}{2} (2 \cdot 10 + (30-1)d)

2400 = 15 (20 + 29d)

160 = 20 + 29d

140 = 29d

d = \frac{140}{29} = 4.83

Answer:

d = 4.83

11. Find the 20th term of the AP: 50, 45, 40, 35, ...

Solution: Using

a_n = a + (n-1)d

$a = 50$ ,
$d = 45 - 50 = -5$ ,
$n = 20$ .

a_{20} = 50 + (20-1)(-5)

= 50 - 95

= -45

Answer:

-45

12. Find the sum of the first 12 terms of the AP: 5, 10, 15, ...

Solution: Using

S_n = \frac{n}{2} (2a + (n-1)d)

$a = 5$ ,
$d = 5$ ,
$n = 12$ .

S_{12} = \frac{12}{2} (2 \cdot 5 + (12-1)5)

= 6 (10 + 55)

= 6 \cdot 65

= 390

Answer:

390

Here are the remaining 13 unique arithmetic progression (AP) problems with step-by-step solutions.

13. The 5th term of an AP is 25, and the 10th term is 50. Find the first term and common difference.

Solution: Using

a_n = a + (n-1)d

For

a_5 = 25

a + 4d = 25

For

a_{10} = 50

a + 9d = 50

Subtracting equations:

(a + 9d) - (a + 4d) = 50 - 25

5d = 25

d = 5

Substituting

d

a + 4(5) = 25

a + 20 = 25

a = 5

Answer:

a = 5

d = 5

14. Find the number of terms in the AP: 7, 14, 21, ..., 133.

Solution: Using

a_n = a + (n-1)d

$a = 7$ ,
$d = 14 - 7 = 7$ ,
$a_n = 133$ .

133 = 7 + (n-1)7

133 - 7 = (n-1)7

126 = (n-1)7

n - 1 = 18

n = 19

Answer:

19

15. The sum of the first 40 terms of an AP is 8200. The first term is 15. Find the common difference.

Solution: Using

S_n = \frac{n}{2} (2a + (n-1)d)

8200 = \frac{40}{2} (2 \cdot 15 + (40-1)d)

8200 = 20 (30 + 39d)

410 = 30 + 39d

380 = 39d

d = \frac{380}{39} = 9.74

Answer:

d = 9.74

16. Find the 25th term of the AP: 1, 4, 7, 10, ...

Solution: Using

a_n = a + (n-1)d

$a = 1$ ,
$d = 3$ ,
$n = 25$ .

a_{25} = 1 + (25-1)3

= 1 + 72

= 73

Answer:

73

17. Find the sum of the first 18 terms of the AP: 10, 20, 30, ...

Solution: Using

S_n = \frac{n}{2} (2a + (n-1)d)

$a = 10$ ,
$d = 10$ ,
$n = 18$ .

S_{18} = \frac{18}{2} (2 \cdot 10 + (18-1)10)

= 9 (20 + 170)

= 9 \cdot 190

= 1710

Answer:

1710

18. The first term of an AP is -3, and the 21st term is 57. Find the common difference.

Solution: Using

a_n = a + (n-1)d

57 = -3 + (21-1)d

57 + 3 = 20d

60 = 20d

d = \frac{60}{20} = 3

Answer:

d = 3

19. The sum of the first 50 terms of an AP is 5050. Find the first term if the common difference is 2.

Solution: Using

S_n = \frac{n}{2} (2a + (n-1)d)

5050 = \frac{50}{2} (2a + (50-1)2)

5050 = 25 (2a + 98)

202 = 2a + 98

2a = 104

a = 52

Answer:

a = 52

20. Find the 15th term of the AP: 100, 97, 94, ...

Solution: Using

a_n = a + (n-1)d

$a = 100$ ,
$d = 97 - 100 = -3$ ,
$n = 15$ .

a_{15} = 100 + (15-1)(-3)

= 100 - 42

= 58

Answer:

58

21. Find the number of terms in the AP: 50, 45, 40, ..., 5.

Solution: Using

a_n = a + (n-1)d

$a = 50$ ,
$d = -5$ ,
$a_n = 5$ .

5 = 50 + (n-1)(-5)

5 - 50 = (n-1)(-5)

-45 = (n-1)(-5)

n-1 = 9

n = 10

Answer:

10

22. The 6th term of an AP is 19, and the 11th term is 34. Find the first term and common difference.

Solution: Using

a_n = a + (n-1)d

For

a_6 = 19

a + 5d = 19

For

a_{11} = 34

a + 10d = 34

Subtracting equations:

(a + 10d) - (a + 5d) = 34 - 19

5d = 15

d = 3

Substituting

d

a + 5(3) = 19

a + 15 = 19

a = 4

Answer:

a = 4

d = 3

23. Find the sum of the first 22 terms of the AP: 8, 14, 20, ...

Solution: Using

S_n = \frac{n}{2} (2a + (n-1)d)

$a = 8$ ,
$d = 6$ ,
$n = 22$ .

S_{22} = \frac{22}{2} (2 \cdot 8 + (22-1)6)

= 11 (16 + 126)

= 11 \cdot 142

= 1562

Answer:

1562

24. The sum of the first 60 terms of an AP is 18300. Find the first term if the common difference is 5.

Solution: Using

S_n = \frac{n}{2} (2a + (n-1)d)

18300 = \frac{60}{2} (2a + (60-1)5)

18300 = 30 (2a + 295)

610 = 2a + 295

2a = 315

a = 157.5

Answer:

a = 157.5

25. Find the 30th term of the AP: -5, -2, 1, ...

Solution: Using

a_n = a + (n-1)d

$a = -5$ ,
$d = 3$ ,
$n = 30$ .

a_{30} = -5 + (30-1)3

= -5 + 87

= 82

Answer:

82

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Calculation problems involving geometric progression

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1. Find the 10th term of the GP: 3, 6, 12, 24, ...

Solution: The formula for the nth term of a GP is:

a_n = a r^{(n-1)}

Where:

$a = 3$ (first term),
$r = \frac{6}{3} = 2$ (common ratio),
$n = 10$ .

a_{10} = 3 \times 2^{(10-1)}

= 3 \times 2^9

= 3 \times 512

= 1536

Answer:

1536

2. Find the sum of the first 6 terms of the GP: 5, 10, 20, 40, ...

Solution: The sum of the first

n

terms of a GP is given by:

S_n = \frac{a (r^n - 1)}{r - 1}, \quad \text{for } r \neq 1

Here,

$a = 5$ ,
$r = 2$ ,
$n = 6$ .

S_6 = \frac{5(2^6 - 1)}{2 - 1}

= 5(64 - 1)

= 5 \times 63

= 315

Answer:

315

3. The 4th term of a GP is 48, and the 7th term is 384. Find the first term and common ratio.

Solution: Using

a_n = a r^{(n-1)}

For

a_4 = 48

a r^3 = 48

For

a_7 = 384

a r^6 = 384

Dividing both equations:

\frac{a r^6}{a r^3} = \frac{384}{48}

r^3 = 8

r = 2

Substituting in

a r^3 = 48

a(2^3) = 48

a(8) = 48

a = 6

Answer:

a = 6

r = 2

4. Find the sum to infinity of the GP: 20, 10, 5, 2.5, ...

Solution: The sum to infinity of a GP is:

S_{\infty} = \frac{a}{1 - r}, \quad \text{for } |r| < 1

Here,

$a = 20$ ,
$r = \frac{10}{20} = 0.5$ .

S_{\infty} = \frac{20}{1 - 0.5}

= \frac{20}{0.5}

= 40

Answer:

40

5. Find the number of terms in the GP: 4, 12, 36, ..., 2916.

Solution: Using

a_n = a r^{(n-1)}

$a = 4$ ,
$r = \frac{12}{4} = 3$ ,
$a_n = 2916$ .

2916 = 4 \times 3^{(n-1)}

Dividing by 4:

\frac{2916}{4} = 3^{(n-1)}

729 = 3^{(n-1)}

Since

729 = 3^6

n - 1 = 6

n = 7

Answer:

7

6. Find the 8th term of the GP: 2, 6, 18, 54, ...

Solution: Using the formula:

a_n = a r^{(n-1)}

Where:

$a = 2$ ,
$r = \frac{6}{2} = 3$ ,
$n = 8$ .

a_8 = 2 \times 3^{(8-1)}

= 2 \times 3^7

= 2 \times 2187

= 4374

Answer:

4374

7. Find the sum of the first 5 terms of the GP: 1, 4, 16, 64, ...

Solution:

S_n = \frac{a (r^n - 1)}{r - 1}

$a = 1$ ,
$r = 4$ ,
$n = 5$ .

S_5 = \frac{1(4^5 - 1)}{4 - 1}

= \frac{1024 - 1}{3}

= \frac{1023}{3} = 341

Answer:

341

8. The 6th term of a GP is 81, and the 10th term is 6561. Find the common ratio.

Solution:

a_6 = a r^5 = 81

a_{10} = a r^9 = 6561

Dividing both:

\frac{a r^9}{a r^5} = \frac{6561}{81}

r^4 = 81

r = 3

Answer:

r = 3

9. Find the sum to infinity of the GP: 8, 4, 2, 1, ...

Solution:

S_{\infty} = \frac{a}{1 - r}

$a = 8$ ,
$r = \frac{4}{8} = 0.5$ .

S_{\infty} = \frac{8}{1 - 0.5}

= \frac{8}{0.5} = 16

Answer:

16

10. Find the number of terms in the GP: 3, 9, 27, ..., 2187.

Solution:

2187 = 3 \times 3^{(n-1)}

\frac{2187}{3} = 3^{(n-1)}

729 = 3^{(n-1)}

n-1 = 6

n = 7

Answer:

7

11. The sum of the first 7 terms of a GP is 5461. The first term is 1, and the common ratio is 2. Find $S_7$ .

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Solution:

S_7 = \frac{1(2^7 - 1)}{2 - 1}

= 2^7 - 1

= 128 - 1

= 127

Answer:

127

12. Find the 15th term of the GP: 5, 10, 20, ...

Solution:

a_{15} = 5 \times 2^{(15-1)}

= 5 \times 2^{14}

= 5 \times 16384

= 81920

Answer:

81920

13. The 9th term of a GP is 512, and the 3rd term is 8. Find the common ratio.

Solution:

\frac{a r^8}{a r^2} = \frac{512}{8}

r^6 = 64

r = 2

Answer:

r = 2

14. Find the sum to infinity of the GP: 6, 2, $\frac{2}{3}$ ,...

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Solution:

S_{\infty} = \frac{a}{1 - r}

$a = 6$ ,
$r = \frac{2}{6} = \frac{1}{3}$ .

S_{\infty} = \frac{6}{1 - \frac{1}{3}}

= \frac{6}{\frac{2}{3}}

= 6 \times \frac{3}{2}

= 9

Answer:

9

15. The sum of the first 8 terms of a GP is 765. The first term is 3, and the common ratio is 2. Find $S_8$ .

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Solution:

S_8 = \frac{3(2^8 - 1)}{2 - 1}

= 3(256 - 1)

= 3 \times 255

= 765

Answer:

765

16. Find the number of terms in the GP: 7, 14, 28, ..., 1792.

Solution:

1792 = 7 \times 2^{(n-1)}

\frac{1792}{7} = 2^{(n-1)}

256 = 2^{(n-1)}

n - 1 = 8

n = 9

Answer:

9

17. Find the sum to infinity of the GP: 12, 4, $\frac{4}{3}$ ,..

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Solution:

S_{\infty} = \frac{a}{1 - r}

$a = 12$ ,
$r = \frac{4}{12} = \frac{1}{3}$ .

S_{\infty} = \frac{12}{1 - \frac{1}{3}}

= \frac{12}{\frac{2}{3}}

= 12 \times \frac{3}{2}

= 18

Answer:

18

18. The sum of the first 10 terms of a GP is 3072. The first term is 1, and the common ratio is 2. Find $S_{10}$ .

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Solution:

S_{10} = \frac{1(2^{10} - 1)}{2 - 1}

= 2^{10} - 1

= 1024 - 1

= 1023

Answer:

1023

19. Find the 20th term of the GP: 10, 5, 2.5, ...

Solution:

a_{20} = 10 \times 0.5^{(20-1)}

= 10 \times 0.5^{19}

= 10 \times \frac{1}{524288}

= \frac{10}{524288}

Answer:

\frac{10}{524288}

20. Find the sum of the first 12 terms of the GP: 1, -1, 1, -1, ...

Solution: Since

r = -1

, the sum alternates.

S_{12} = \frac{1( (-1)^{12} - 1 )}{-1 - 1}

= \frac{(1 - 1)}{-2}

= 0

Answer:

0

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Calculation problem involving sum of the nth term of a geometric progression

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1. Find the sum of the first 7 terms of the GP: 2, 6, 18, 54, ...

Solution: The sum of the first

n

terms of a GP is given by:

S_n = \frac{a (r^n - 1)}{r - 1}, \quad \text{for } r \neq 1

Here:

$a = 2$ ,
$r = \frac{6}{2} = 3$ ,
$n = 7$ .

S_7 = \frac{2(3^7 - 1)}{3 - 1}

= \frac{2(2187 - 1)}{2}

= 2(1093)

= 2186

Answer:

2186

2. Find the sum of the first 6 terms of the GP: 4, -12, 36, -108, ...

Solution: Using:

S_n = \frac{a (r^n - 1)}{r - 1}

Here:

$a = 4$ ,
$r = -3$ ,
$n = 6$ .

S_6 = \frac{4((-3)^6 - 1)}{-3 - 1}

= \frac{4(729 - 1)}{-4}

= \frac{4(728)}{-4}

= -728

Answer:

-728

3. The sum of the first 8 terms of a GP is 3280. The first term is 5, and the common ratio is 2. Find $S_8$ .

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Solution: Using:

S_n = \frac{a (r^n - 1)}{r - 1}

$a = 5$ ,
$r = 2$ ,
$n = 8$ .

S_8 = \frac{5(2^8 - 1)}{2 - 1}

= 5(256 - 1)

= 5(255)

= 1275

Answer:

1275

4. Find the sum of the first 10 terms of the GP: 1, 3, 9, ...

Solution: Using:

S_n = \frac{a (r^n - 1)}{r - 1}

$a = 1$ ,
$r = 3$ ,
$n = 10$ .

S_{10} = \frac{1(3^{10} - 1)}{3 - 1}

= \frac{3^{10} - 1}{2}

= \frac{59049 - 1}{2}

= \frac{59048}{2}

= 29524

Answer:

29524

5. Find the sum of the first 9 terms of the GP: 7, 21, 63, ...

Solution: Using:

S_n = \frac{a (r^n - 1)}{r - 1}

$a = 7$ ,
$r = 3$ ,
$n = 9$ .

S_9 = \frac{7(3^9 - 1)}{3 - 1}

= \frac{7(19683 - 1)}{2}

= \frac{7(19682)}{2}

= 68887

Answer:

68887

6. The sum of the first 5 terms of a GP is 242. The first term is 2, and the common ratio is 3. Find $S_5$ .

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Solution: Using:

S_n = \frac{a (r^n - 1)}{r - 1}

$a = 2$ ,
$r = 3$ ,
$n = 5$ .

S_5 = \frac{2(3^5 - 1)}{3 - 1}

= \frac{2(243 - 1)}{2}

= 2(121)

= 242

Answer:

242

7. Find the sum of the first 7 terms of the GP: 5, 10, 20, 40, ...

Solution: Using:

S_n = \frac{a (r^n - 1)}{r - 1}

$a = 5$ ,
$r = 2$ ,
$n = 7$ .

S_7 = \frac{5(2^7 - 1)}{2 - 1}

= 5(128 - 1)

= 5(127)

= 635

Answer:

635

8. Find the sum of the first 12 terms of the GP: 3, 6, 12, ...

Solution: Using:

S_n = \frac{a (r^n - 1)}{r - 1}

$a = 3$ ,
$r = 2$ ,
$n = 12$ .

S_{12} = \frac{3(2^{12} - 1)}{2 - 1}

= 3(4096 - 1)

= 3(4095)

= 12285

Answer:

12285

9. Find the sum of the first 6 terms of the GP: 10, -20, 40, -80, ...

Solution: Using:

S_n = \frac{a (r^n - 1)}{r - 1}

$a = 10$ ,
$r = -2$ ,
$n = 6$ .

S_6 = \frac{10((-2)^6 - 1)}{-2 - 1}

= \frac{10(64 - 1)}{-3}

= \frac{10(63)}{-3}

= -210

Answer:

-210

10. The sum of the first 4 terms of a GP is 1875. The first term is 5, and the common ratio is 5. Find $S_4$ .

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Solution: Using:

S_n = \frac{a (r^n - 1)}{r - 1}

$a = 5$ ,
$r = 5$ ,
$n = 4$ .

S_4 = \frac{5(5^4 - 1)}{5 - 1}

= \frac{5(625 - 1)}{4}

= \frac{5(624)}{4}

= \frac{3120}{4}

= 780

Answer:

780

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Progression | Jamb Mathematics

Calculation problem involving arithmetic progression

1. Find the 10th term of the AP: 2, 5, 8, 11, ...

2. Find the sum of the first 15 terms of the AP: 3, 7, 11, 15, ...

3. The 7th term of an AP is 20, and the 13th term is 50. Find the first term and common difference.

4. Find the number of terms in the AP: 4, 10, 16, ..., 88.

5. The sum of the first 20 terms of an AP is 1000. The first term is 7. Find the common difference.

6. Find the 12th term of the AP: 6, 13, 20, 27, ...

7. Find the sum of the first 25 terms of the AP: 2, 8, 14, 20, ...

8. Find the number of terms in the AP: 3, 7, 11, ..., 87.

9. The 8th term of an AP is 36, and the 16th term is 84. Find the first term and common difference.

10. The sum of the first 30 terms of an AP is 2400. The first term is 10. Find the common difference.

11. Find the 20th term of the AP: 50, 45, 40, 35, ...

12. Find the sum of the first 12 terms of the AP: 5, 10, 15, ...

13. The 5th term of an AP is 25, and the 10th term is 50. Find the first term and common difference.

14. Find the number of terms in the AP: 7, 14, 21, ..., 133.

15. The sum of the first 40 terms of an AP is 8200. The first term is 15. Find the common difference.

16. Find the 25th term of the AP: 1, 4, 7, 10, ...

17. Find the sum of the first 18 terms of the AP: 10, 20, 30, ...

18. The first term of an AP is -3, and the 21st term is 57. Find the common difference.

19. The sum of the first 50 terms of an AP is 5050. Find the first term if the common difference is 2.

20. Find the 15th term of the AP: 100, 97, 94, ...

21. Find the number of terms in the AP: 50, 45, 40, ..., 5.

22. The 6th term of an AP is 19, and the 11th term is 34. Find the first term and common difference.

23. Find the sum of the first 22 terms of the AP: 8, 14, 20, ...

24. The sum of the first 60 terms of an AP is 18300. Find the first term if the common difference is 5.

25. Find the 30th term of the AP: -5, -2, 1, ...

Calculation problems involving geometric progression

1. Find the 10th term of the GP: 3, 6, 12, 24, ...

2. Find the sum of the first 6 terms of the GP: 5, 10, 20, 40, ...

3. The 4th term of a GP is 48, and the 7th term is 384. Find the first term and common ratio.

4. Find the sum to infinity of the GP: 20, 10, 5, 2.5, ...

5. Find the number of terms in the GP: 4, 12, 36, ..., 2916.

6. Find the 8th term of the GP: 2, 6, 18, 54, ...

7. Find the sum of the first 5 terms of the GP: 1, 4, 16, 64, ...

8. The 6th term of a GP is 81, and the 10th term is 6561. Find the common ratio.

9. Find the sum to infinity of the GP: 8, 4, 2, 1, ...

10. Find the number of terms in the GP: 3, 9, 27, ..., 2187.

12. Find the 15th term of the GP: 5, 10, 20, ...

13. The 9th term of a GP is 512, and the 3rd term is 8. Find the common ratio.

16. Find the number of terms in the GP: 7, 14, 28, ..., 1792.

19. Find the 20th term of the GP: 10, 5, 2.5, ...

20. Find the sum of the first 12 terms of the GP: 1, -1, 1, -1, ...

Calculation problem involving sum of the nth term of a geometric progression

1. Find the sum of the first 7 terms of the GP: 2, 6, 18, 54, ...

2. Find the sum of the first 6 terms of the GP: 4, -12, 36, -108, ...

4. Find the sum of the first 10 terms of the GP: 1, 3, 9, ...

5. Find the sum of the first 9 terms of the GP: 7, 21, 63, ...

7. Find the sum of the first 7 terms of the GP: 5, 10, 20, 40, ...

8. Find the sum of the first 12 terms of the GP: 3, 6, 12, ...

9. Find the sum of the first 6 terms of the GP: 10, -20, 40, -80, ...

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