Question

1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4...... In the above sequence what is the number of the position 2888 of the sequence.

Answer 1

Answer:

1

Step-by-step explanation:

1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4...... In the above sequence what is the number of the position 2888 of the sequence.

1st Sequence  = 1 + 2 + 3 + 4 = 10 Digits

2nd Sequence = 2 + 4 + 6 + 8 = 20 Digits

3rd Sequence = 3 + 6 + 9 + 12 = 30 Digits

10 + 20 + 30 +......................................

a = 10  d = 10

Sum of n Sequences ≥ 2888

(n/2)(10 + (n-1)10) ≥ 2888

=>5n²  ≥ 2888

=> n > 24

For n = 24

Sum of Sequences = (24/2)(10 + 23*10) = 2880

2888 - 2880 =  8

For n = 25

1  will be 25 times  from 2881 to 3006

=> number position of 2888​ is  1

Answer 2

Answer:

1

Step-by-step explanation: