Question

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

• A. $43,45,....,75$
• B. $43,45,47,....,79$
• C. $43,45,....,85$
• D. $43,45,...,89$

Answer 1

The correct option is D $43,45,...,89$

Let $n$ consecutive odd integers are $2m+1,2m+3,...,2m+2n-1$
Now according to problem
$(2m+1)+(2m+3)+....+(2m+2n-1)$
$={45}^2-{21}^2$
$\Rightarrow 2m+\underset{\left(ntimes\right)}{2m+.....}+2m+1+3+....+(2n-1)$
$={45}^2-{21}^2$
$\Rightarrow 2mn+n^2={45}^2-{21}^2$
$\Rightarrow (m+n{)}^2-m^2={45}^2-{21}^2$
$\Rightarrow (n+m)=45,m=21\Rightarrow n=24$
$\Rightarrow$ Consecutive odd integers are $43,45,47,...,89$
Hence choice (d) is correct answer.