Let k the smallest integer such that 7k is the smallest multiple
of 7 such that the sum of the series

7k+(7k+7)+(7k+14)+ ∙∙∙ (to 50 terms) = 12075

Th formula for the sum of an arithmetic series is

, where n=50, a1=7k, d=7







And since S50=12075



Divide both sides by 25





 

The smallest integer = 7k = 7(10) = 70
The largest integer is the 50th term











Answers: smallest = 70, biggest = 413

Edwin

Use the formula for the sum of the first 50 terms of an arithmetic progression with the common difference 7,

. = 12075.     (1)


Since  = , you can rewrite (1) in the form


. = 12075,   or

. = 24150,

 =   ====>   =  = 70.


Thus you found the smallest term of the AP. It is 70.


Then the biggest term is 70 + 7*49 = 413.


Happily, the first term is multiple of 7 and the common difference is 7, so all the terms of the progression 
are consecutive integers multiple of 7.