Question

Find the sum of two consecutive positive integers, such that the sum of their squares is 365.

• A. 27

Answer 1

The correct option is A 27

Let the numbers be $x\&x+1.$

Given, the sum of squares of the positive consecutive integers is 365.
$\therefore x^2+(x+1{)}^2=365$

Lets solve
$x^2+x^2+2x+1=3652x^2+2x-364=0x^2+x-182=0$
(Dividing by 2)

Factorise,
$x^2+14x-13x-182=0x(x+14)-13(x+14)=0(x+14)(x-13)=0\therefore x=13\&-14$

Since numbers are positive integers,
hence $x=13$.

So, required consecutive numbers are $13\&14.$
Sum of the consecutive numbers = 13 + 14 = 27