Challenge!

1) Since p, q and r are in A.P., we have r - q = q - p and hence:
aRr-q = aRq-p for any real values of a and R
which comes to aRr / aRq = aRq / aRp → aRr-1 / aRq-1 = aRq-1 / aRp-1, implying that aRp-1, aRq-1 and aRr-1 are in G.P.
Therefore the pth, qth and rth terms of any G.P. are in G.P.
2) Let me assume that your question does not have any typo mistake. For such question, we have to solve it by means of Newton's-Raphson method which is a method of approximation of root as follows:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Maths/Newtonmethod.jpg

From the graph above, we can see that the general expression of each iterative value of root is:
xn+1 = xn - [F(xn)/F'(xn)]
Usually the iteration can top if |xn+1 - xn| < 10-4.
For the estimation of x0, find it such that F(x0) is reasonably small.
Now, let F(x) = 2x+2 - 6x - 2×32x+2, then
F(x) = 4×2x - 6x - 2×9×32x
= 4×2x - 6x - 18×9x
Therefore F'(x) = (4ln 2)2x - (ln 6)6x - (18ln 9)9x by means of logarithmic differentiation.
So for solving F(x) = 0, we start with x0 = -1 that F(x0) = -1/6
So using the above theory, we can perform the iteration as follows:
x1 = -1.0615
x2 = -1.0526
x3 = -1.0548
x4 = -1.0543
x5 = -1.0544
x6 = -1.0544
Therefore, we can see that the root can be approximated to x = -1.0544 (Corr. to 4 d.p.).

If p,q, r are in A.P., show that the pth, qth and rth terms of any G.P. are in G.P.
If p, q and r are in A.P.,
Let the common difference be d, so
q = p + d ... (1)
r = q + d = p + 2d ... (2)
Consider a G.P. with first term a and common ratio R,
For the n-th term = S(n)
The p-th term = S(p)
= aRp-1
The q-th term = S(q)
= aRq-1
= aRp+d-1 【From (1)】 ... (3)
The r-th term = S(r)
= aRr-1
= aRp+2d-1 【From (2)】 ... (4)
Consider the S(r)/S(q)
S(r)/S(q) = aRp+2d-1 / aRp+d-1
S(r)/S(q) = R(p+2d-1)-(p+d-1)
S(r)/S(q) = Rd .... (5)
Consider the S(q)/S(p)
S(q)/S(p) = aRp+d-1 / aRp-1
S(q)/S(p) = R(p+d-1)-(p-1)
S(q)/S(p) = Rd .... (6)
From (5), (6), as the ratio is the same, so
the p-th term, q-th term and r-th term of any G.P. are in G.P.


2007-01-31 13:50:59 補充：
回應你的補充：你的問題應該有少少錯誤，應為 2^(2x+2)2. Solve for x. 2^(2x+2) - 6^x - (2)3^(2x+2) = 0 (2²)(2^2x) - [(2)(3)]^x - (2)(3^2x)(3²) = 04(2^x)² - (2^x)(3^x) - 18(3^x)² = 018(3^x)² + (2^x)(3^x) - 4(2^x)² = 0[9(3^x) - 4(2^x)][2(3^x) + 1] = 0續..

2007-01-31 13:54:00 補充：
9(3^x) - 4(2^x) = 0 or 2(3^x) + 1 = 0(3^2)(3^x) - (2^2)(2^x) = 0 or 3^x = -1/2 (rejected as 3^x ＞ 0)3^(x+2) - 2^(x+2) = 03^(x+2) = 2^(x+2)log[3^(x+2)] = log[2^(x+2)](x+2)log3 = (x+2)log2(x+2)(log3 - log2) = 0x+2 = 0 【As log3-log2 &gt; 0】x = -2

2007-01-31 13:54:33 補充：
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