omckyyo的部落格

標題:

~Arithmetic sequence~

發問:

In an as,the sum of three consecutive terms is 24 and the product of these terms is 312. Find two possible sets of values for these terms.

最佳解答:

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let a be the first term, d be the common difference first term = a second term = a+d third term = a+2d a+(a+d)+(a+2d)=24 3a+3d=24 3(a+d)=24 a+d=8 a=8-d -----(1) a(a+d)(a+2d)=312 (8-d)[(8-d)+d][(8-d)+2d]=312 (8-d)(8)(8+d)=312 8(64-d^2)=312 512-8d^2=312 200-8d^2=0 8d^2=200 d^2=25 d= 5 or -5 a= 3 or 13 respectively first set: 3, 8, 13 second set: 13, 8, 5 first set: 2007-11-12 15:01:33 補充： second set: 13, 8, 3

其他解答:

Let the 3 consecutive terms be a-d, a and a+d where a is a constant and d is the common difference. Then, (a-d) + a + (a+d) = 24 or 3a = 24 => a = 8 -----------(1) and (a-d)a(a+d) = 312 => a(a^2-d^2) = 312 => 8(8^2 - d^2) = 312 => d^2 = 25 => d = 5 or -5 Therefore the possible terms can be (3, 8, 13) or (13, 8, 3)7638E748EED250B0