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506
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avatar+4568 

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
 #1
avatar+111331 
+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)

 

 

cool cool cool

 Jan 23, 2018
edited by CPhill  Jan 24, 2018

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