# Math help!!!?

Given the arithmetic sequence a,x,b and given a^2 + b^2 = 148 determine a, x, and b. I know the answer but I do not know how to solve algebraically.

### 1 Answer

- MathmomLv 73 years agoFavorite Answer
There is no way to determine this.

Since a, x, b are in AP, then

x − a = b − x

2x = a + b

x = (a + b) / 2

a² + b² = 148 has infinitely many real solutions (although only a few integer solutions)

So given any values for a and b we can find a unique value for x ((a+b)/2) that will make a, x, b an arithmetic progression.

So we don't have enough information to find unique values for a, x, b.

——————————————————————————————

As mentioned above, a² + b² = 148 has only a few integer solutions:

a = 2, b = 12 ----> x = (2+12)/2 = 7

a = 2, b = −12 ----> x = (2−12)/2 = −5

a = −2, b = 12 ----> x = (−2+12)/2 = 5

a = −2, b = −12 ----> x = (−2−12)/2 = −7

a = 12, b = 2 ----> x = (12+2)/2 = 7

a = 12, b = −2 ----> x = (12−2)/2 = 5

a = −12, b = 2 ----> x = (−12+2)/2 = −5

a = −12, b = −2 ----> x = (−12−2)/2 = −7

So there are 8 integer solutions for a, x, b

2, 7, 12

2, −5, −12

−2, 5, 12

−2, −7, −12

12, 7, 2

12, 5, −2

−12, −5, 2

−12, −7, −2

All of the above are arithmetic sequences with a² + b² = 148

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