Question

Sum of certain consecutive odd positive integers is ${57}^2-{13}^2$ . Find the first integer of that series .

• A. 29
• B. 27
• C. 25
• D. 31

Answer 1

The correct option is B 27

Let the odd integers be $2m+1,2m+3,2m+5$, ... and let their number be $n$. Then
${57}^2-{13}^2=\left[n/2\right][2(2m+1)+(n-1)2]=n(2m+n)=2mn+n^2\Rightarrow {57}^2-{13}^2={[n+m]}^2-m^2$
Hence , $m=13$ and $n+m=57$ or $n=57-13=44$
Therefore , the required odd integers are $27,29,31,...,113$