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What number comes next in each sequence below? |
1. 1, 3, 6, 10, 15, 21, 28, _______
Answer: 361; 1 + 2 = 3; 1 + 2 + 3 = 6; 1 + 2 + 3 + 4 = 10; 1 + 2 + 3 + 4 + 5 = 15; 1 + 2 + 3 + 4 + 5 + 6 = 21; 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28; 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36; The nth term in the sequence is given by n (n + 1)/2, and the numbers are often referred to as triangular numbers. 2. 1, 3, 7, 15, 31, 63, 127, _______
Answer: 25521 - 1 = 2 - 1 = 1; 22 - 1 = 4 - 1 = 3; 23 - 1 = 8 - 1 = 7; 24 - 1 = 16 - 1 = 15; 25 - 1 = 32 - 1 = 31; 26 - 1 = 64 - 1 = 63; 27 - 1 = 128 - 1 = 127; 28 - 1 = 256 - 1 = 255; The nth term in the sequence is given by 2n - 1. So, the difference between two consecutive numbers forms the following simple sequence: 2, 4, 8, 16, 32, 64, 128, ... 3. 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, _______
Answer: 550 + 1 = 1; 1 + 1 = 2; 1 + 2 = 3; 2 + 3 = 5; 3 + 5 = 8; 5 + 8 = 13; 8 + 13 = 21; 13 + 21 = 34; 21 + 34 = 55; Each term (starting with the third term) in the sequence is the sum of the two terms preceding it. The series is often referred to as the Fibonacci series. Fibonacci (1175) believed that this series was followed by various natural phenomena. In fact, the number of leaves on the stems of particular plants follows this series.
4. 15, 12, 24, 20, 33, 28, 42, _______
Answer: 36The odd terms of the sequence continually increase by 9, i.e., 15 + 9 = 24; 24 + 9 = 33 ; 33 + 9 = 42; ... The even terms of the sequence continually increase by 8, i.e., 12 + 8 = 20; 20 + 8 = 28; 28 + 8 = 36; ... 5. 2, 6, 12, 20, 30, 42, 56, _______
Answer: 72(1)(2) = 2; (2)(3) = 6; (3)(4) = 12; (4)(5) = 20; (5)(6) = 30; (6)(7) = 42; (7)(8) = 56; (8)(9) = 72; The nth term in the sequence is given by n (n + 1). 6. 6, 24, 60, 120, _______, 336, 504, 720
Answer: 210(1)(2)(3) = 6; (2)(3)(4) = 24; (3)(4)(5) = 60; (4)(5)(6) = 120; (5)(6)(7) = 210; (6)(7)(8) = 336; (7)(8)(9) = 504; (8)(9)(10) = 720; The nth term in the sequence is given by n (n + 1) (n + 2). 7. 1, 2, 6, 24, 120, _______
Answer: 7201 = 1; (1)(2) = 2; (1)(2)(3) = 6; (1)(2)(3)(4) = 24; (1)(2)(3)(4)(5) = 120; (1)(2)(3)(4)(5)(6) = 720; The nth term in the sequence is given by n! (factorial of n), which is defined as the product of all integers from 1 to n. 8. 0, 1, 2, 7, 20, 61, 182, _______
Answer: 5473(0) + 2(1) = 2; 3(1) + 2(2) = 7; 3(2) + 2(7) = 20; 3(7) + 2(20) = 61; 3(20) + 2(61) = 182; 3(61) + 2(182) = 547; The nth term in the sequence is given by tn = 3 tn - 2 + 2 tn - 1. A term (starting with the third) in the sequence is a linear combination of the preceding two terms. So, let the nth term in the sequence be given by tn = a tn - 2 + b tn - 1. For n = 3, 2 = a(0) + b(1) For n = 4, 7 = a(1) + b(2) Thus, a = 3 and b = 2. 9. 1/4, 0, 1, -3, 13, -51, 205, _______
Answer: -8194(1/4) - 3(0) = 1; 4(0) - 3(1) = -3; 4(1) - 3(-3) = 13; 4(-3) - 3(13) = -51; 4(13) - 3(-51) = 205; 4(-51) - 3(205) =-819; The nth term in the sequence is given by tn = 4 tn - 2 - 3 tn - 1. A term (starting with the third) in the sequence is a linear combination of the preceding two terms. So, let the nth term in the sequence be given by tn = a tn - 2 + b tn - 1. For n = 3, 1 = a(1/4) + b(0) For n = 4, -3 = a(0) + b(1) Thus, a = 4 and b = -3. 10. 1, 10, 11, 100, 101, 110, 111, _______
Answer: 10001 (base 2) = 1(1) = 1 (base 10); 10 (base 2) = 1(2) + 0(1) = 2 (base 10); 11 (base 2) = 1(2) + 1(1) = 3 (base 10); 100 (base 2) = 1(4) + 0(2) + 0(1) = 4 (base 10); 101 (base 2) = 1(4) + 0(2) + 1(1) = 5 (base 10); 110 (base 2) = 1(4) + 1(2) + 0(1) = 6 (base 10); 111 (base 2) = 1(4) + 1(2) + 1(1) = 7 (base 10); 1000 (base 2) = 1(8) + 0(4) + 0(2) + 0(1) = 8 (base 10); The sequence is simply 1, 2, 3, 4, 5, ... in the binary (base 2) system. 11. 1, 2, 10, 37, 101, _______
Answer: 2262 - 1 = 1; 10 - 2 = 8; 37 - 10 = 27; 101 - 37 = 64; The differences between two consecutive numbers are 1, 8, 27, 64, ... (cubes of integers starting with 1). So, 101 + 53 = 101 + 125 = 226 12. 2, 5, 10, 17, _______, 37, 50, 65
Answer: 26The terms are merely one more than the squares of integers starting with 1. Thus, 12 + 1 = 1 + 1 = 2; 22 + 1 = 4 + 1 = 5; 32 + 1 = 9 + 1 = 10; 42 + 1 = 16 + 1 = 17; 52 + 1 = 25 + 1 = 26; 62 + 1 = 36 + 1 = 37; Alternatively, the differences between consecutive terms form the following simple sequence: 3, 5, 7, 9, 11, 13, 15, ... 13. 7, 26, 63, 124, _______, 342
Answer: 215The terms are merely one less than the cubes of integers starting with 2. Thus, 23 - 1 = 8 - 1 = 7; 33 - 1 = 27 - 1 = 26; 43 - 1 = 64 - 1 = 63; 53 - 1 = 125 - 1 = 124; 63 - 1 = 216 - 1 = 215; 73 - 1 = 343 - 1 = 342; 14. 2, 12, 36, 80, _______, 252, 392, 576
Answer: 150The terms are the sum of the squares and cubes of integers, starting with 1. Thus, 12 + 13 = 1 + 1 = 2; 22 + 23 = 4 + 8 = 12; 32 + 33 = 9 + 27 = 36; 42 + 43 = 16 + 64 = 80; 52 + 53 = 25 + 125 = 150; 62 + 63 = 36 + 216 = 252; 72 + 73 = 49 + 343 = 392; 82 + 83 = 64 + 512 = 576; 15. 2, 5, 17, 65, _______, 1025
Answer: 257The terms are merely one more than the powers of 4. Thus, 40 + 1 = 1 + 1 = 2; 41 + 1 = 4 + 1 = 5; 42 + 1 = 16 + 1 = 17; 43 + 1 = 64 + 1 = 65; 44 + 1 = 256 + 1 = 257; 45 + 1= 1024 + 1 = 1025; 16. 9, 729, 8, _______, 7, 343
Answer: 512The odd terms are merely the integers starting with 9 in descending order. The even terms are the cubes of the odd terms. Thus, 93 = 9 x 9 x 9 = 729; 83 = 8 x 8 x 8 = 512; 73 = 7 x 7 x 7 = 343; 17. 361, 289, _______, 169, 121
Answer: 225The terms are merely the squares of odd integers starting with 19 in descending order. Thus, 192 = 19 x 19 = 361; 172 = 17 x 17 = 289; 152 = 15 x 15 = 225; 132 = 13 x 13 = 169; 112 = 11 x 11 = 121; The differences between two consecutive numbers are 72, 64, 56, 48, ... (a simple sequence starting with 72 and continually decreasing by 8). 18. 1/12, 1/2, 9/8, 13/6, 17/4, _______
Answer: 21/2The numerators (starting with 1) increase continually by 4, and the denominators (starting with 12) decrease continually by 2. The numerators are 1, 5, 9, 13, 17, 21. The denominators are 12, 10, 8, 6, 4, 2. So, the fractions are 1/12, 5/10, 9/8, 13/6, 17/4, and 21/2. Note that 5/10 is equivalent to 1/2.
Try the Quiz : IQ Tests : Number Sequences II
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