Question

write three arithmetic sequence with 99 as the sum of the three consecutive terms.

Answer 1

First sequence is x + x+1 + x+2 = 99

3x+3=99

x+1=33

x=32

Then, sequence is 32, 33, 34

Second sequence is x + x-1 + x-2 = 99

3x-3=99

x-1=33

x=34

Then, sequence is 34, 33, 32

Third sequence is x-1 + x-2 + x-3 = 99

3x-6=99

x-2 = 33

x=35

Then, the sequence is 34, 33, 32

Since they're the consecutive terms. Nothing else can be possible.

Answer 2

first x+1,x+2,x+3=99.