Question

If $A$ denotes the property that two elements of $A=\{1,5,9,13,...,1093\}$ add up to $1094,$ then the maximum number of elements in $A$ that show this property can be

• A. $126$
• B. $136$
• C. $137$
• D. $138$

Answer 1

The correct option is C $137$

Since the elements in A are in AP, therefore the number of elements in $A=\frac{1093-1}{4}+1=274\dots (n=\frac{t_n-a}{d}+1)$

Since the sum of equidistant terms in AP is equal to the sum of first and the last terms$=1+1093=1094$

So the maximum number of elements that add up to $1094=\frac{274}{2}=137$