Geometric Progression — Quick Slides (1–10)
1. If \(5,\;x,\;y,\;z,\;405\) are the first five terms of a G.P., determine \(x,y,z\).
Let common ratio be \(r\). Terms: \(5,5r,5r^{2},5r^{3},5r^{4}=405\).
So \(5r^{4}=405\Rightarrow r^{4}=81=3^{4}\). Hence \(r=\pm3\).
If \(r=3\): \(x=15,\ y=45,\ z=135\). If \(r=-3\): \(x=-15,\ y=45,\ z=-135\).
\(\boxed{(x,y,z)=(15,45,135)\ \text{or}\ (-15,45,-135)}\)
2. The second term of a G.P. is \(b\) and the common ratio is \(r\). If the product of first three terms equals \(64\), find \(b\).
Terms: \(a,\ ar,\ ar^{2}\) with \(ar=b\). Product \(=a^{3}r^{3}\).
Since \(a=\dfrac{b}{r}\), product \(=(\tfrac{b}{r})^{3}r^{3}=b^{3}\).
Therefore \(b^{3}=64\Rightarrow b=4\) (real root).
\(\boxed{b=4}\)
3(i). For the G.P. \(2,6,18,54,\dots\), find \(T_{10}\) and \(T_{p}\).
\(a=2,\ r=3,\ T_{n}=ar^{\,n-1}=2\cdot3^{\,n-1}.\)
Hence \(T_{10}=2\cdot3^{9}\) and \(T_{p}=2\cdot3^{\,p-1}\).
\(\boxed{T_{10}=2\cdot3^{9},\quad T_{p}=2\cdot3^{\,p-1}}\)
3(ii). For the G.P. \(4,-8,16,-32,\dots\), find \(T_{9}\) and \(T_{q}\).
\(a=4,\ r=-2,\ T_{n}=4(-2)^{\,n-1}.\)
So \(T_{9}=4(-2)^{8}=1024\) and \(T_{q}=4(-2)^{\,q-1}\).
\(\boxed{T_{9}=1024,\quad T_{q}=4(-2)^{\,q-1}}\)
3(iii). For the G.P. \(\{\sqrt{5},1,\tfrac{1}{\sqrt{5}},\tfrac{1}{5},\dots\}\), determine \(T_{14}\) and \(T_{n}\).
\(a=\sqrt{5},\ r=\dfrac{1}{\sqrt{5}}\). \(T_{n}=\sqrt{5}\big(\tfrac{1}{\sqrt{5}}\big)^{n-1}\).
Thus \(T_{14}=\sqrt{5}\big(\tfrac{1}{\sqrt{5}}\big)^{13}=5^{-6}=\dfrac{1}{15625}\).
\(\boxed{T_{14}=\dfrac{1}{15625},\quad T_{n}=\sqrt{5}\big(\tfrac{1}{\sqrt{5}}\big)^{n-1}}\)
4(i). If the 3rd term of a G.P. equals the square of the 1st term and the 5th term equals \(729\), determine the G.P.
Let \(a,ar,ar^{2},ar^{3},ar^{4}\). Given \(ar^{2}=a^{2}\Rightarrow r^{2}=a\).
Also \(ar^{4}=729\). With \(a=r^{2}\), \(r^{6}=729=3^{6}\Rightarrow r=\pm3\).
Hence \(a=r^{2}=9\) and G.P. is \(9,27,81,\dots\) or \(9,-27,81,\dots\).
\(\boxed{9,27,81,\dots\ \text{or}\ 9,-27,81,\dots}\)
4(ii). If \(T_{p+q}=a\) and \(T_{p-q}=b\) for a G.P., find \(T_{p}\).
Let \(T_{n}=Ar^{\,n-1}\). Then \(ab=A^{2}r^{2p-2}=(Ar^{p-1})^{2}=T_{p}^{2}\).
Therefore \(T_{p}=\pm\sqrt{ab}\) (take \(+\sqrt{ab}\) for positive terms).
\(\boxed{T_{p}=\sqrt{ab}}\)
4(iii). If the \(n\)th term equals \(p\), show the product of first \(2n-1\) terms equals \(p^{2n-1}\).
\(T_{n}=p=Ar^{n-1}\Rightarrow A=pr^{1-n}\).
Product \(P=A^{2n-1}r^{\sum_{0}^{2n-2}k}=A^{2n-1}r^{(2n-1)(n-1)}\).
Substitute \(A\) to get \(P=p^{2n-1}\).
\(\boxed{P=p^{2n-1}}\)
5(i). Locate \(243\sqrt{2}\) in the G.P. \(\{\sqrt{2},\sqrt{6},3\sqrt{2},3\sqrt{6},\dots\}\).
\(a=\sqrt{2},\ r=\sqrt{3},\ T_{n}=\sqrt{2}\,3^{\tfrac{n-1}{2}}\).
Solve \(3^{\tfrac{n-1}{2}}=3^{5}\Rightarrow n=11\).
\(\boxed{11^{\text{th}}\ \text{term}}\)
5(ii). Is \(256\) a term of the G.P. \(3,6,12,24,\dots\)?
\(T_{n}=3\cdot2^{n-1}\). If \(3\cdot2^{n-1}=256\) then \(2^{n-1}=\dfrac{256}{3}\), not a power of \(2\).
Thus \(256\) is not a term.
\(\boxed{\text{No}}\)
