Find five consecutive integers whose sum is 195.

n ; n + 1 ; n + 2 ; n + 3 ; n + 4 − f i v e co n sec u t i v e in t e g ers ( n ) + ( n + 1 ) + ( n + 2 ) + ( n + 3 ) + ( n + 4 ) = 195 + n + 1 + n + 2 + n + 3 + n + 4 = 195 5 n + 10 = 195 ∣ s u b t r a c t 10 f ro m b o t h s i d es 5 n = 185 ∣ d i v i d e b o t h s i d es b y 5 = 37 A n s w er : 37 ; 38 ; 39 ; 40 ; 41 ​

Answered by Anonymous | 2024-06-10

To do this, come up with three numbers. These are n, n+1, n+2, n+3, and n+4.
To solve, you do this:
n + n + 1 + n + 2 + n + 3 + n + 4 = 195 5 n + 10 = 195 5 n + ( 10 − 10 ) = ( 195 − 10 ) 5 n = 185 5 5 n ​ = 5 185 ​ n = 37
Then, substitute 37 into the numbers:
n=37 n+1=37+1=38 n+2=37+2=39 n+3=37+3=40 n+4=37+4=41
The five consecutive integers are 37, 38, 39, 40, and 41.

Answered by Anonymous | 2024-06-10

The five consecutive integers that sum up to 195 are 37, 38, 39, 40, and 41. This was calculated by defining the first integer as n and solving the resulting equation. The integers can be expressed as n , n + 1 , n + 2 , n + 3 , n + 4 .
;

Answered by Anonymous | 2024-10-14