The mean of seven positive integers is 16. When the smallest of these seven integers is removed, the sum of the remaining six integers is 108. What is the value of the integer that was removed?

Answer 1

[ n+ (n+2) + (n+4) + .....(n+12) ] /7 = 16

[ (n+2) + (n+4) + .... (n+12) ] = 108
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well, multiply the first one by 7
[ n+ (n+2) + (n+4) + .....(n+12) ] = 16*7 = 112
[ (n+2) + (n+4) + .... (n+12) ] = 108
------------------------------------------------subtract
n = 112 - 108

Answer 2

I read even numbers so increased by 2 each time. Just do increase by 1
same system, same answer

Answer 3

Thanks

Answer 4

You are welcome.

Answer 5

Doesn't say anything about the numbers being in an arithmetic sequence

total of the seven numbers = 6(16) = 112
total of six numbers after one was removed = 108
number removed must have been 4

Answer 6

You guys smell like poo

Answer 7

No Arithmetic Sequence Involved In Solving Process

mean of seven positive integers = 16

Total Sum of Seven Numbers- 7 (16) = 112
Small Integer Removed - 108 (Mean)
112 - 108 = 4
The difference is equivalent to 4.