Sum of 3 numbers in AP is 21 and their product is 231. Find the numbers. This problem is to be solved by using high school math only. The answer is that numbers are 3,7,11. The solution includes assuming that the required numbers be a, a-d, a+d . But i am not able to understand why we are assuming these , since its not given that the numbers is consecutive in the AP, so all we know that they can be of form a, a+nd, a+md . A similar question which says that numbers are consecutive in the given AP has a similar assumption that numbers are a, a-d, a+d. but why can't the numbers be a , a+d , or a+2d?
 A: Let the numbers be $a-d,a$ and $a+d$, clearly $d$ is the common difference.
So, $a-d+a+a+d=21\implies 3a=21\implies a=7$
and $a(a-d)(a+d)=231\implies 7(7^2-d^2)=231$
$\implies 7^2-d^2=33\implies d^2=16\implies d=\pm4$
So, the numbers are $7,7\pm4$ i.e, $3,7,11$

If we take the numbers to be $a,a+d,a+2d$ where $d$ is the common difference,
$a+a+d+a+2d=21\implies a+d=7,a=7-d$
Now, $a(a+d)(a+2d)=231\implies (7-d)7(7+d)=231\implies 7^2-d^2=33\implies d=\pm4$
A: The numbers can be $a, a+d, a+2d$.  Instead, they chose to start counting at term $-1$ instead of $0$.  This made it easier to solve the simultaneous equations because the sum becomes $a-d+a+a+d=3a$ and you get $a=7$ immediately.
If we want to use $a,a+d,a+2d$ we get $3a+3d=21$ or $a+d=7$  Then from the product we have $a(a+d)(a+2d)=7a(7+d)=231,\ \  a(7+d)=33,\ \ (7-d)(7+d)=33=49-d^2$ which you can solve  to get $d=4, a=3$ or $d=-4,a=11$
A: We are assuming (a-d), a , (a+d) to simple our problem. Since we know that those numbers which are consecutive are in A.P. so we choose it that way. We can choose a,a+d, a+2d, but that will just lead us to a big calculation. We can solve it like this
Let numbers be a,a+d, a+2d
now, a+(a+d)+(a+2d)= 21
     3a+3d=21
     a+d=7
Again. a.(a+d).(a+2d)= 231
       a.7.(7+d)=231
       (7-d).(7+d)=33
        49-d^2=33
        d=4 or -4
and we get a= 3 or 11
and thus we get the numbers as 3,7,11.
