The sum of four numbers in an AP is 0 and their product is 9. Find the numbers.

Dear student
Let a be the first term in the series and l be the last term in the series and d be the common difference,
There are 4 terms in the series so n=4.

Therefore l = a₄  = a + (n-1)d
                           =  a + (4-1)d
                           =  a + 3d

Sum of first n terms in an A.P,

S_n = n/2 [2a + (n-1)d]
     = n/2 [a + a + (n-1)d] 

But a + (n-1)d = l

Therefore,
S_n = n/2 [a + l]

It's given that S₄  = 0,

So, 
S₄ = 4/2[a + l] = 0
a + l = 0
a = -l

But l = a + 3d

​Hence,
a = -a - 3d
2a = -3d
d = -2a/3

Now that we have got a relation between a and d,

The numbers of the series are respectively,
a, a+d, a+2d, a+3d

​It's given that a(a+d)(a+2d)(a+3d) = 9

Substituting the value for d above,

a(a - 2a/3)(a - 4a/3)(a - 6a/3) = 9
 
a(a/3)(-a/3)(-3a/3) = 9

a(a/3)(a/3)(a) = 9

a⁴/9 = 9

a⁴ =  81

a⁴ = 3⁴

a = ±3 and d = ∓2.

Therefore, the numbers are
                        3,1,-1,-3  
                            OR
                        -3,-1,1,3                
Regards