Find 4 numbers i arithmetic progression

Question

Four integers in A.P their sum is 24 and their product is 945 .Find them I assumed them as $$a-d,a,a+d,a+2d$$ From the givens we get $$4a+2d=24$$ then $$2a+d=12$$ Then apply the product to get 4th degree equation my question is there a simpler way to avoid the the 4th degree equation?

Answer 1

we have the equations $$4a+6d=24$$ and $$a(a+d)(a+2d)(a+3d)=945$$ from the first equation we get $$d=4-\frac{2}{3}a$$ plugging this in the second equation and factorizing:$$(a-3)(a-9)(a^2-12a-315)=0$$ you will have to solve the equation $$a^4+24a^3-144a^2+3456a-8505=0$$ often is it useful to make the ansatz $$a^4+24a^3-144a^2+3456a-8505=(x^2+Ax+B)(x^2+Cx+D)$$

Answer 2

Note that the product of integers $$a-d,a,a+d,a+2d$$ should be 945.

Upon factorization of $$945 = 3\times 5\times 7\times 9$$ we find out that our terms are $$ 3,5,7,9$$

That is $a=5$ and $d=2$