41 should be in between 14 and 122
41.
Let's see. 122 + 14 = 136. 136 + 14 = 150. 150 + 14 = ?
71
This is a number sequence in which each term is found by multiplying the previous term by three and then subtracting one. So the next term would be (122 times 3) minus 1which equals 366 - 1 = 365
365.
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.
Let's see. 122 + 14 = 136. 136 + 14 = 150. 150 + 14 = ?
71
2,5,14 - - 122 Difference between 5 & 2 = 3 = 3^(1) Difference between 14 & 5 = 9 = 3^(2) This leads one to suppose the next difference will be 3^(3) = 27 Hence 14 + 27 = 41 Similarly the next difference again will 3^(4) = 81 Hence 41 + 81 = 122 As required. So the sequence is 2,5,14,41,122. NB The next difference again is 3^(5) = 243 Hence 122 + 243 = 365 is the next term in the sequence. =
To determine the missing numbers in the sequence 70 -- 84, we need to identify the pattern between the two known numbers. The difference between 70 and 84 is 14. To find the missing numbers, we need to continue the pattern of adding 14 to the previous number. Therefore, the missing numbers in the sequence are 98 and 112.
This is a number sequence in which each term is found by multiplying the previous term by three and then subtracting one. So the next term would be (122 times 3) minus 1which equals 366 - 1 = 365
2 (+3) 5 (+9) 14 (+27) 41 (+81) 122 in this series... the difference is multiplied by 3. like 3,9,27,81,243...
14
119. un = (-n4 + 15n3 - 77n2 + 225n + 192)/6
41 122 365