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# A question!

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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?

Jan 23, 2018

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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)   Jan 23, 2018
edited by CPhill  Jan 24, 2018