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if a,b,c are in A.P. then prove that ac - b² = (-a² -c² + 2ac )/4?
回答 (2)
2015-02-22 1:57 pm
a,b,c are in AP
→(a+c)/2=b
Squaring both sides gives
(a²+c²+2ac)/4=b²
Hence ac−b²=ac −(a²+c²+2ac)/4
=( 4ac−a²−b²−2ac)/4
=( 2ac−a²−b²)/4(−a²−b²+2ac)/4
→(a+c)/2=b
Squaring both sides gives
(a²+c²+2ac)/4=b²
Hence ac−b²=ac −(a²+c²+2ac)/4
=( 4ac−a²−b²−2ac)/4
=( 2ac−a²−b²)/4(−a²−b²+2ac)/4
2015-02-22 1:35 pm
Assuming a, b, and c are consecutive elements of the sequence,
b = a + d for some d
c = a + 2d for some d
ac - b^2 =
a(a + 2d) - (a + d)^2 =
a^2 + 2ad - a^2 - 2ad - d^2 =
-d^2
(-a^2 - c^2 + 2ac)/4 =
(-a^2 - (a + 2d)^2 + 2a(a + 2d))/4 =
(-a^2 - a^2 - 4ad - 4d^2 + 2a^2 + 4ad)/4 =
-4d^2/4 =
-d^2
Therefore, ac - b^2 = (-a^2 - c^2 + 2ac)/4
b = a + d for some d
c = a + 2d for some d
ac - b^2 =
a(a + 2d) - (a + d)^2 =
a^2 + 2ad - a^2 - 2ad - d^2 =
-d^2
(-a^2 - c^2 + 2ac)/4 =
(-a^2 - (a + 2d)^2 + 2a(a + 2d))/4 =
(-a^2 - a^2 - 4ad - 4d^2 + 2a^2 + 4ad)/4 =
-4d^2/4 =
-d^2
Therefore, ac - b^2 = (-a^2 - c^2 + 2ac)/4
收錄日期: 2021-04-21 01:21:22
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