An algebra problem by Akash Hossain
Algebra
Level
3
Consider the set of numbers that follows an arithmetic progression, .
Find the minimum number of elements in this set such that their sum is equal to 126.
The answer is 2.
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1 solution
It's simple.
In the S= 3 , 8 , 1 3 , . . . . . . , 1 1 8
Given that ( m = l a r g e s t n u m b e r i n t h e ( S ) s e t . ) a n d i n t h e ( S ) s e t l a r g e s t n u m b e r i s ( 1 1 8 ) . 1 2 6 − ( m = 1 1 8 ) = 8
Thus 1 2 8 = 1 1 8 + 8
∴ the answer is only two elements .