Having trouble with G.P. Pls help.?

2016-06-24 10:55 am
If p, q, r, s are in G.P. then proce that p^2-q^2, q^2-r^2, r^2-s^2 are also in G.P.

回答 (2)

2016-06-24 12:03 pm
Call the common ratio k.

Then q = pk, r = pk^2, and s = pk^3

So now we can rewrite:

p^2 - q^2 = p^2 - p^2k^2 = p^2(1 - k^2)

q^2 - r^2 = p^2k^2 - p^2k^4 = p^2k^2(1 - k^2)

r^2 - s^2 = p^2k^4 - p^2k^6 = p^2k^4(1 - k^2)

This is a g.p., with common ratio k^2
2016-06-24 11:09 am
Let R be the common ratio of the G.P. p, q, r, s.
Then q = pR, r = pR² and s = pR³

[q² - r²] / [p² - q²]
= [(pR)² - (pR²)²] / [p² - (pR)²]
= [p²R² - p²R⁴] / [p² - p²R²]
= [p²R²(1 - R²)] / [p²(1 - R²)]
= R²

[r² - s²] = [q² - r²]
= [(pR²)² - (pR³)²] / [(pR)² - (pR²)²]
= [p²R⁴ - p²R⁶] / [p²R² - p²R⁴]
= [p²R⁴(1 - R²)] / p²R²[p²(1 - R²)]
= R²

Since [q² - r²] / [p² - q²] = [r² - s²] = [q² - r²],
then p² - q², q² - r², r² - s² are also in G.P.


收錄日期: 2021-04-18 15:14:59
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160624025548AAri9f7

檢視 Wayback Machine 備份