An algebra problem by Zulqarnain Ansari

$\displaystyle {\text{Let, required integers be }} n, n + 1, n + 2, n + 3 {\text{ and }} n + 4.\\ {\text{So, according to question,}} \\ \Rightarrow \dfrac{n + n + 1 + n +2 + n + 3 + n + 4}{5} = 12 \\ \implies 5n + 10 = 60 \\ \implies 5n = 50 \implies n = 10 \\ \therefore {\text{Least integer}} \rightarrow \boxed{10} \\ {\text{Greatest integer}} \rightarrow \boxed{14} \\ \\ Sum = 10 + 14 = \color{#CEBB00}{\boxed{24}}$

Since there is x-2, x-1, x, x+1, x+2 the sum of the biggest and smallest integers is 2x and x=12 so 2x=24

let the middle number x 5x=60 x=12 so 10+14=24