Question

Three consecutive odd natural numbers are such that the product of the first and third exceeds four times the middle by $1$, then the numbers are $3,5$ and $7$.

• A. Yes
• B. No
• C. Ambiguous
• D. Data insufficient

Answer 1

The correct option is A Yes

Let the numbers be $2n-1,2n+1$ and $2n+3$
Then,
$(2n-1)(2n+3)-4(2n+1)=1$
$4n^2-2n+6n-3-8n-4=1$
$4n^2-4n-8=0$
$n^2-n-2=0$
$n^2-2n+n-2=0$
$(n-2)(n+1)=0$
$n=-1$ or $2$
Neglect the negative value, $n=2$
Hence, the numbers are. $3,5,7$