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

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 = a

_{4}

_{ }= 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 S

_{4}

_{ }= 0,

So,

S

_{4}= 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

a

^{4}/9 = 9

a

^{4}= 81

a

^{4}= 3

^{4}

**a =**±3 and d =

**∓**2.

Therefore, the numbers are

3,1,-1,-3

OR

-3,-1,1,3

Regards

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