41 should be in between 14 and 122
The pattern in the sequence is that each number is obtained by multiplying the previous number by 3 and subtracting 1. 1 x 3 - 1 = 2 2 x 3 - 1 = 5 5 x 3 - 1 = 14 14 x 3 - 1 = 41 41 x 3 - 1 = 122 Therefore, the next number in the sequence would be 122 x 3 - 1 = 365.
The next number in the sequence is 15 - the sequence goes x2, +2, -1
The simplest sequence appears to be 15 1*2=2 2+2=4 4-1=3 3*2=6 6+2=8 8-1=7 7*2=14 14+2=16 16-1=15 Multiply by 2 Add 2 Subtract 1 Repeat
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The sequence has the formula is 3x - 1 2 → 6 - 1 = 5 5 →15 - 1 = 14 14 →42 -1 = 41 41 → 123 - 1 = 122 The missing number is therefore 41.
Type your answer here... The sequence has the formula is 3x - 1 2 → 6 - 1 = 5 5 →15 - 1 = 14 14 →42 -1 = 41 41 → 123 - 1 = 122 The missing number is therefore 41.
41 122 365
41 should be in between 14 and 122
2 (+3) 5 (+9) 14 (+27) 41 (+81) 122 in this series... the difference is multiplied by 3. like 3,9,27,81,243...
The pattern in the sequence is that each number is obtained by multiplying the previous number by 3 and subtracting 1. 1 x 3 - 1 = 2 2 x 3 - 1 = 5 5 x 3 - 1 = 14 14 x 3 - 1 = 41 41 x 3 - 1 = 122 Therefore, the next number in the sequence would be 122 x 3 - 1 = 365.
15 The sequence seems to be x2, +2, -1.
The next number in the sequence is 15 - the sequence goes x2, +2, -1
Add 3n-1. For example 1 + 30 = 2 2 + 31 = 5 5 + 32 = 14 14 + 33 = 41
The simplest sequence appears to be 15 1*2=2 2+2=4 4-1=3 3*2=6 6+2=8 8-1=7 7*2=14 14+2=16 16-1=15 Multiply by 2 Add 2 Subtract 1 Repeat
The simplest sequence appears to be 15 1*2=2 2+2=4 4-1=3 3*2=6 6+2=8 8-1=7 7*2=14 14+2=16 16-1=15 Multiply by 2 Add 2 Subtract 1 Repeat
x 2 + 2 - 1 is the repeating sequence 1 x 2 = 2 2 + 2 = 4 4 - 1 = 3 3 x 2 = 6 6 + 2 = 8 8 - 1 = 7 7 x 2 = 14 14 + 2 = 16 16 - 1 = 15 the next number is 15