A sequence of numbers begin 1, 1, 1, 2, 2, 3 and repeats this pattern for ever. What is the sum of `141^(st), 143^(rd), and 145^(th)` number?,

Options

A. 6
B. 7
C. 5
D. 4

Answer

D

Explanation

The given sequence of numbers follows the pattern:
`{:(111/(3.s),22/(2.s),3/(1.s)):}`
So, it has a cycle of length 6.
So, `141 = 6 xx 23+3 `
So, `141^(st)` number = 1
`143^(rd)` number = 2`
`145^(th)` number = 1
Therefore, Required sum
`= 1+ 2+1 =4`