Geometric Progression

Geometric Progression — Very Short Answer Type (Problems 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: \(5,\;5r,\;5r^{2},\;5r^{3},\;5r^{4}\).

Given \(5r^{4}=405\Rightarrow r^{4}=81=3^{4}\).

So \(r=\pm3\).

If \(r=3\): \(x=5r=15,\ y=5r^{2}=45,\ z=5r^{3}=135\).

If \(r=-3\): \(x=-15,\ y=45,\ z=-135\).

Answer: \(\boxed{(15,45,135)\ \text{or}\ (-15,45,-135)}\)

2. The second term of a G.P. is \(b\) and the common ratio is \(r\). If the product of the first three terms equals \(64\), find \(b\).

First three terms: \(a,\ ar,\ ar^{2}\).

Given second term \(ar=b\) ⇒ \(a=\dfrac{b}{r}\).

Product \(=a\cdot ar\cdot ar^{2}=a^{3}r^{3}\).

Substitute \(a=\dfrac{b}{r}\): product \(=(\tfrac{b}{r})^{3}r^{3}=b^{3}\).

So \(b^{3}=64\Rightarrow b=4\) (real root).

Answer: \(\boxed{b=4}\)

3(i). For the G.P. \(2,6,18,54,\dots\), find \(T_{10}\) and \(T_{p}\).

First term \(a=2\).

Common ratio \(r=3\).

General term \(T_{n}=ar^{n-1}=2\cdot3^{n-1}\).

Hence \(T_{10}=2\cdot3^{9}\).

And \(T_{p}=2\cdot3^{p-1}\).

Answer: \(\boxed{T_{10}=2\cdot3^{9},\ T_{p}=2\cdot3^{p-1}}\)

3(ii). For the G.P. \(4,-8,16,-32,\dots\), find \(T_{9}\) and \(T_{q}\).

First term \(a=4\).

Ratio \(r=-2\).

General term \(T_{n}=4(-2)^{n-1}\).

So \(T_{9}=4(-2)^{8}=1024\).

And \(T_{q}=4(-2)^{q-1}\).

Answer: \(\boxed{T_{9}=1024,\ T_{q}=4(-2)^{q-1}}\)

3(iii). For the G.P. \(\{\sqrt{5},1,\tfrac{1}{\sqrt{5}},\tfrac{1}{5},\dots\}\), determine \(T_{14}\) and \(T_{n}\).

First term \(a=\sqrt{5}\).

Ratio \(r=\dfrac{1}{\sqrt{5}}\).

General term \(T_{n}=\sqrt{5}\big(\tfrac{1}{\sqrt{5}}\big)^{n-1}\).

Compute \(T_{14}=\sqrt{5}\big(\tfrac{1}{\sqrt{5}}\big)^{13}=5^{-6}=\dfrac{1}{15625}\).

Answer: \(\boxed{T_{14}=\dfrac{1}{15625},\ T_{n}=\sqrt{5}\big(\tfrac{1}{\sqrt{5}}\big)^{n-1}}\)

4(i). If the 3rd term equals the square of the 1st and the 5th term equals \(729\), determine the G.P.

Let terms be \(a,\ ar,\ ar^{2},\ ar^{3},\ ar^{4}\).

Given \(ar^{2}=a^{2}\Rightarrow r^{2}=a\) (assume \(a\ne0\)).

Given \(ar^{4}=729\); with \(a=r^{2}\) ⇒ \(r^{6}=729=3^{6}\).

So \(r=\pm3\) and \(a=r^{2}=9\).

Hence G.P. is \(9,27,81,\dots\) (if \(r=3\)) or \(9,-27,81,\dots\) (if \(r=-3\)).

Answer: \(\boxed{9,27,81,\dots\ \text{or}\ 9,-27,81,\dots}\)

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 \(T_{p+q}=Ar^{p+q-1}=a\) and \(T_{p-q}=Ar^{p-q-1}=b\).

Multiply: \(ab=A^{2}r^{2p-2}=(Ar^{p-1})^{2}=T_{p}^{2}\).

Therefore \(T_{p}=\pm\sqrt{ab}\) (choose + when terms positive).

Answer: \(\boxed{T_{p}=\sqrt{ab}\ \text{(or }-\sqrt{ab}\text{)}}\)

4(iii). If the n-th term equals p, show the product of the first 2n-1 terms equals \(p^{2n-1}\).

Given \(T_{n}=p=Ar^{n-1}\Rightarrow A=pr^{1-n}.\)

Product \(P=\prod_{k=1}^{2n-1}Ar^{k-1}=A^{2n-1} r^{\sum_{k=0}^{2n-2}k}.\)

Sum \(\sum_{k=0}^{2n-2}k=(2n-1)(n-1).\)

Substitute \(A\): \(P=(pr^{1-n})^{2n-1} r^{(2n-1)(n-1)}=p^{2n-1}.\)

Result: \(\boxed{P=p^{2n-1}}\)

5(i). Locate \(243\sqrt{2}\) in the G.P. \(\{\sqrt{2},\sqrt{6},3\sqrt{2},3\sqrt{6},\dots\}\).

First term \(a=\sqrt{2}\).

Common ratio \(r=\dfrac{\sqrt{6}}{\sqrt{2}}=\sqrt{3}\).

General term \(T_{n}=\sqrt{2}(\sqrt{3})^{n-1}=\sqrt{2}\,3^{(n-1)/2}.\)

Solve \(\sqrt{2}\,3^{(n-1)/2}=243\sqrt{2}\Rightarrow 3^{(n-1)/2}=3^{5}.\)

So \((n-1)/2=5\Rightarrow n=11.\)

Answer: \(\boxed{243\sqrt{2}\ \text{is the 11th term}}\)

5(ii). Is \(256\) a term of the G.P. \(3,6,12,24,\dots\)?

First term \(a=3\), ratio \(r=2\).

General term \(T_{n}=3\cdot2^{n-1}\).

If \(256\) were a term, \(2^{n-1}=\dfrac{256}{3}\), not a power of 2.

Therefore \(256\) is not a term.

Answer: \(\boxed{\text{No}}\)

6(i). Complete the progression: \(\dfrac{11}{4},\ \_,\ \_,\ \_,\ \dfrac{22}{27}\).

Let common ratio be \(r\), and \(a=\dfrac{11}{4}\).

Given \(a r^{4}=\dfrac{22}{27}\Rightarrow r^{4}=\dfrac{\tfrac{22}{27}}{\tfrac{11}{4}}=\dfrac{8}{27}\).

So \(r=(\tfrac{8}{27})^{1/4}=(\tfrac{2}{3})^{3/4}\).

Missing terms: \(\dfrac{11}{4}r,\ \dfrac{11}{4}r^{2},\ \dfrac{11}{4}r^{3}\).

Answer: \(\boxed{\dfrac{11}{4}r,\ \dfrac{11}{4}r^{2},\ \dfrac{11}{4}r^{3}}\) with \(r=(\tfrac{2}{3})^{3/4}\)

6(ii). Complete the progression: \(-1,\ \_,\ \_,\ \_,\ -16,\dots\).

Let \(a=-1\). Given \(ar^{4}=-16\Rightarrow r^{4}=16\).

So \(r=\pm2\).

If \(r=2\): sequence \(-1,-2,-4,-8,-16\).

If \(r=-2\): sequence \(-1,2,-4,8,-16\).

Answer: \(\boxed{-1,-2,-4,-8,-16\ \text{or}\ -1,2,-4,8,-16}\)

6(iii). Complete the progression: \(0.1,\ \_,\ \_,\ \_,\ 0.000032,\dots\).

First term \(a=0.1=10^{-1}\), last \(ar^{4}=0.000032=32\times10^{-6}\).

So \(r^{4}=\dfrac{32\times10^{-6}}{10^{-1}}=32\times10^{-5}\).

Hence \(r=\big(32\times10^{-5}\big)^{1/4}\).

Missing terms: \(0.1r,\ 0.1r^{2},\ 0.1r^{3}\) (exact root form given).

Answer: \(\boxed{0.1r,\ 0.1r^{2},\ 0.1r^{3}}\) where \(r=\big(32\times10^{-5}\big)^{1/4}\)

7(i). If \(3r+1,\ 7r,\ 10r+8\) are in G.P., determine \(r\).

Middle term squared equals product of extremes: \((7r)^{2}=(3r+1)(10r+8)\).

So \(49r^{2}=30r^{2}+34r+8\Rightarrow 19r^{2}-34r-8=0.\)

Solve quadratic: \(r=\dfrac{34\pm\sqrt{34^{2}+608}}{38}=\dfrac{34\pm42}{38}\).

Thus \(r=2\) or \(r=-\tfrac{4}{19}\).

Answer: \(\boxed{r=2\ \text{or}\ r=-\tfrac{4}{19}}\)

7(ii). If \(2r,\ 4r+1,\ 6r+2\) are in G.P., determine \(r\).

\((4r+1)^{2}=(2r)(6r+2)=12r^{2}+4r\).

So \(16r^{2}+8r+1=12r^{2}+4r\Rightarrow 4r^{2}+4r+1=0\).

Discriminant \(=16-16=0\) ⇒ double root \(r=-\tfrac{1}{2}\).

Answer: \(\boxed{r=-\tfrac{1}{2}}\)

8. Insert four geometric means between \(\dfrac{7}{4}\) and \(56\).

Let \(a=\dfrac{7}{4}\). With four means, \(ar^{5}=56\).

So \(r^{5}=\dfrac{56}{7/4}=32\Rightarrow r=2.\)

Four means: \(ar=\dfrac{7}{2},\ ar^{2}=7,\ ar^{3}=14,\ ar^{4}=28.\)

Answer: \(\boxed{\dfrac{7}{2},\ 7,\ 14,\ 28}\)

9. If positive unequal numbers \(a,b,c\) are in G.P., prove \(a+c>2b\).

Write \(a=pr^{-1},\ b=p,\ c=pr\) with \(r>0\) and \(r\ne1\).

Then \(a+c=p\big(r+\tfrac{1}{r}\big)\).

Since \(r+\tfrac{1}{r}\ge2\) with equality only at \(r=1\), and \(r\ne1\), we have \(r+\tfrac{1}{r}>2\).

Hence \(a+c>2p=2b\).

Result: \(\boxed{a+c>2b}\)

10. Can three distinct numbers be simultaneously in A.P. and G.P.? Justify.

Suppose \(a,b,c\) are in A.P. and G.P.; then \(2b=a+c\) and \(b^{2}=ac\).

From \(2b=a+c\): \((a+c)^{2}=4b^{2}=4ac\).

So \(a^{2}-2ac+c^{2}=0\Rightarrow (a-c)^{2}=0\).

Thus \(a=c\), contradicting distinctness; hence impossible unless \(a=b=c\).

Answer: \(\boxed{\text{Not possible unless }a=b=c}\)