Question

The consecutive odd integers whose sum is $45^2-21^2$ are

• A. $43,45,.....,75$
• B. $43,45,.....,79$
• C. $43,45,.....,85$
• D. $43,45,.....,89$

Answer 1

Answer: (D)

Let $n$ consecutive odd integers are $2m+1,2m+3,2m+5,\dots .,2m+2n-1$
Given that, $45^2-21^2=(2m+1)+(2m+3)+(2m+5)+\dots \dots +(2m+2n-1)$
$=2mn+(1+3+5+\dots \dots +(2n-1))$
$=2mn+n^2$
$=m^2+2mn+n^2-m^2$
$\Rightarrow 45^2-21^2=(m+n{)}^2-m^2$
$\Rightarrow m+n=45$ and $m=21$
$\Rightarrow n=24$ and $m=21$
Hence, the numbers are
$43,45,\dots ,89$