Math

The sum of 11 consecutive positive integers is 2002. What is the greatest of these 11 integers?
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  1. a 1 = first number

    a 2 = a 1 + 1 = second number

    a 3 = a 1 + 2 = third number

    a 4 = a 1 + 3 = fourth number

    a 5 = a 1 + 4 = fifth number

    a 6 = a 1 + 5 = sixth number

    a 7 = a 1 + 6 = seventh number

    a 8 = a 1 + 7 = eighth numer

    a 9 = a 1 + 8 = ninth number

    a 10 = a 1 + 9 = tenth number

    a 11 = a 1 + 10 = eleventh number

    a 1 + a 2 + a 3 + a 4 + a 5 + a 6 + a 7 + a 8 + a 9 + a 10 + a 11 = 2002

    a 1 + a 1 + 1 + a 1 + 2 + a 1 + 3 + a 1 + 4 + a 1 + 5 + a 1 + 6 + a 1 + 7 + a 1 + 8 + a 1 + 9 + a 1 + 10 = 2002

    11 a 1 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 2002

    11 a 1 + 55 = 2002

    11 a 1 = 2002 - 55

    11 a 1 = 1947 Divide both sides by 11

    a 1 = 1947 / 11

    a 1 = 177

    a2 = a 1 + 1 = 177 + 1 = 178

    a 3 = a 1 + 2 = 177 + 2 = 179

    a 4 = a 1 + 3 = 177 + 3 = 180

    a 5 = a 1 + 4 = 177 + 5 = 181

    a 6 = a 1 + 5 = 177 + 5 = 182

    a 7 = a 1 + 6 = 177 + 6 = 183

    a 8 = a 1 + 7 = 177 + 74 = 184

    a 9 = a 1 + 8 = 177 + 8 = 185

    a 10 = a 1 + 9 = 177 + 9 = 186

    a 11 = a 1 + 10 = 177 + 10 = 187

    a 1 + a 2 + a 3 + a 4 + a 5 + a 6 + a 7 + a 8 + a 9 + a 10 + a 11 =

    177 + 178 + 179 + 180 + 181 + 182 + 183 + 184 + 185 + 186 + 187 = 2002

    The greatest is eleventh number:

    a 11 = 187

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  2. Wow, this helped me sooooo much, thank you!!! Keep making posts like this.

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  3. Wrong answer

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