Question

The sum of three consecutive positive integers is 192. Find the integers

Answer 1

Here is your answer ✌

Let the starting integer be n

n+(n+1)+(n+2)=192

3n+3=192

3n=189

n=63

So your integers are:

63+64+65=192

Hope my answer helps.✌

Thank you☺

Be Brainly✌✌

Answer 2

Hey! ! !

Solution :-

☆ Let the integers be "n"

So that,
the consecutive integers be

n, n + 1 , n + 2

:- A/q

Sum of consecutive integers = 192

=> n +( n +1 )+ ( n + 2 ) = 192

=>3n + 3 = 192

=> 3n = 192 - 3 = 189

=> n = 189/3

:- n = 63

So the required no = 63 , 64 , 65

Verify :-

☆ 63 + 64 + 65 = 192

=> 192 = 192

=> LHS = RHS

☆ ☆ ☆ Hop its helpful ☆ ☆ ☆

☆ Regards :- ♡♡《 Nitish kr singh 》♡♡