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