In each blank below a single digit is inserted such that the following six three-digit numbers, in this order, form an arithmetic sequence:

\( 1 \,\underline{\hspace{9pt}}\, \,\underline{\hspace{9pt}}\, \;\;,\;\; \,\underline{\hspace{9pt}}\, \,\underline{\hspace{9pt}}\, 9 \;\;,\;\; 2 \,\underline{\hspace{9pt}}\, 2 \;\;,\;\; \,\underline{\hspace{9pt}}\, 6 \,\underline{\hspace{9pt}}\, \;\;,\;\; 2 \,\underline{\hspace{9pt}}\, \,\underline{\hspace{9pt}}\, \;\;,\;\; \,\underline{\hspace{9pt}}\, 3 \,\underline{\hspace{9pt}} \)

What is the value of the next number in the sequence?

tertre Jan 23, 2018

#1**+1 **

1 _ _ , _ _9, 2 _ 2 , _6_ , 2_ _ , _3_

The last number of the common difference must end in 3

And this must be a 2 digit number

So....the first number must end in 6

The the 4th number must be 265

So..... the 5th number must end in 8

And the last digit of the last term must end in 1 [ and begin with 3 ]

Putting all this together we have that

265 + 2d = 331

2d = 66 ⇒ d = 33

So the first term must be 265 - 3 (66) = 166

So....the series appears to be 166, 199, 232, 265, 298, 331

And the generating formula for the nth term is 166 + 33 (n - 1)

CPhill Jan 23, 2018