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.

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  • 3 years ago
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    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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