Question

find an A. P whose sum is three time of the sum of squares of its terms.

Answer 1

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.