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Question Number 12304 by Gaurav3651 last updated on 18/Apr/17
How many geometric progressions is/are possible contauning 27,8 and 12 as three of its/their terms? (a) 1 (b) 2 (c) 4 (d) infinitely many
$${How}\:{many}\:{geometric}\:{progressions} \\ $$$${is}/{are}\:{possible}\:{contauning}\:\mathrm{27},\mathrm{8} \\ $$$${and}\:\mathrm{12}\:{as}\:{three}\:{of}\:{its}/{their}\:{terms}? \\ $$$$\left({a}\right)\:\:\mathrm{1} \\ $$$$\left({b}\right)\:\:\mathrm{2} \\ $$$$\left({c}\right)\:\:\mathrm{4} \\ $$$$\left({d}\right)\:\:{infinitely}\:{many} \\ $$$$ \\ $$
Answered by mrW1 last updated on 18/Apr/17
let′s assume 8, 12 and 27 are 3 terms of a GP with the common ratio q: 8...(n−1 terms)...12....(m−1 terms)...27 ((27)/(12))=((3×3×3)/(3×2×2))=((3×3)/(2×2))=q^m (i) ((12)/8)=((2×2×3)/(2×2×2))=(3/2)=q^n (ii) q^(m−n) =(3/2) (i)/(ii) ⇒ q=((3/2))^(1/(m−n)) q^n =((3/2))^(n/(m−n)) =(3/2) from (ii) ⇒ (n/(m−n))=1 n=m−n ⇒ m=2n ⇒ q=((3/2))^(1/n) from (ii) i.e. for any n≥1 we can always find a GP whose 1st term is 8, (1+n)−th term is 12 and (1+3n)−th term is 27, and their common ratio is q=((3/2))^(1/n) : T_1 =8 ..... (n−1 terms between) T_(1+n) =8×q^n =8×((3/2))^(n/n) =12 ..... (2n−1 terms between) T_(1+3n) =8×q^(3n) =8×((3/2))^((3n)/n) =8×((3/2))^3 =27 since any n≥1 fulfills this, we have infinitely many such GP. ⇒ Answer (d) is correct. Examples n=1: 8,12,18,27..... n=2: 8,8×((3/2))^(1/2) ,12,12×((3/2))^(1/2) ,18,18×((3/2))^(1/2) ,27.... n=3: 8,8×((3/2))^(1/3) ,8×((3/2))^(2/3) ,12,12×((3/2))^(1/3) ,12×((3/2))^(2/3) ,18,18×((3/2))^(1/3) ,18×((3/2))^(2/3) ,27....
$${let}'{s}\:{assume}\:\mathrm{8},\:\mathrm{12}\:{and}\:\mathrm{27}\:{are}\:\mathrm{3}\:{terms}\:{of} \\ $$$${a}\:{GP}\:{with}\:{the}\:{common}\:{ratio}\:{q}: \\ $$$$\mathrm{8}…\left({n}−\mathrm{1}\:{terms}\right)…\mathrm{12}….\left({m}−\mathrm{1}\:{terms}\right)…\mathrm{27} \\ $$$$ \\ $$$$\frac{\mathrm{27}}{\mathrm{12}}=\frac{\mathrm{3}×\mathrm{3}×\mathrm{3}}{\mathrm{3}×\mathrm{2}×\mathrm{2}}=\frac{\mathrm{3}×\mathrm{3}}{\mathrm{2}×\mathrm{2}}={q}^{{m}} \:\:\:\:\:\:\:\:\left({i}\right) \\ $$$$\frac{\mathrm{12}}{\mathrm{8}}=\frac{\mathrm{2}×\mathrm{2}×\mathrm{3}}{\mathrm{2}×\mathrm{2}×\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}}={q}^{{n}} \:\:\:\:\:\:\:\:\:\:\:\:\:\left({ii}\right) \\ $$$${q}^{{m}−{n}} =\frac{\mathrm{3}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\left({i}\right)/\left({ii}\right) \\ $$$$\Rightarrow\:{q}=\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{{m}−{n}}} \\ $$$${q}^{{n}} =\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\frac{{n}}{{m}−{n}}} =\frac{\mathrm{3}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:{from}\:\left({ii}\right) \\ $$$$\Rightarrow\:\frac{{n}}{{m}−{n}}=\mathrm{1} \\ $$$${n}={m}−{n} \\ $$$$\Rightarrow\:{m}=\mathrm{2}{n} \\ $$$$\Rightarrow\:{q}=\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{{n}}} \:\:\:\:\:\:{from}\:\left({ii}\right) \\ $$$${i}.{e}.\:\:{for}\:{any}\:{n}\geqslant\mathrm{1}\:{we}\:{can}\:{always}\:{find} \\ $$$${a}\:{GP}\:\:{whose}\:\mathrm{1}{st}\:{term}\:{is}\:\mathrm{8},\:\left(\mathrm{1}+{n}\right)−{th}\: \\ $$$${term}\:{is}\:\mathrm{12}\:{and}\:\left(\mathrm{1}+\mathrm{3}{n}\right)−{th}\:{term}\:{is}\:\mathrm{27}, \\ $$$${and}\:{their}\:{common}\:{ratio}\:{is}\:{q}=\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{{n}}} : \\ $$$${T}_{\mathrm{1}} =\mathrm{8} \\ $$$$…..\:\left({n}−\mathrm{1}\:{terms}\:{between}\right) \\ $$$${T}_{\mathrm{1}+{n}} =\mathrm{8}×{q}^{{n}} =\mathrm{8}×\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\frac{{n}}{{n}}} =\mathrm{12} \\ $$$$…..\:\left(\mathrm{2}{n}−\mathrm{1}\:{terms}\:{between}\right) \\ $$$${T}_{\mathrm{1}+\mathrm{3}{n}} =\mathrm{8}×{q}^{\mathrm{3}{n}} =\mathrm{8}×\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\frac{\mathrm{3}{n}}{{n}}} =\mathrm{8}×\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{3}} =\mathrm{27}\: \\ $$$$ \\ $$$${since}\:{any}\:{n}\geqslant\mathrm{1}\:{fulfills}\:{this},\:{we}\:{have} \\ $$$${infinitely}\:{many}\:{such}\:{GP}. \\ $$$$ \\ $$$$\Rightarrow\:{Answer}\:\left({d}\right)\:{is}\:{correct}. \\ $$$$ \\ $$$${Examples} \\ $$$${n}=\mathrm{1}: \\ $$$$\mathrm{8},\mathrm{12},\mathrm{18},\mathrm{27}….. \\ $$$$ \\ $$$${n}=\mathrm{2}: \\ $$$$\mathrm{8},\mathrm{8}×\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} ,\mathrm{12},\mathrm{12}×\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} ,\mathrm{18},\mathrm{18}×\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} ,\mathrm{27}…. \\ $$$$ \\ $$$${n}=\mathrm{3}: \\ $$$$\mathrm{8},\mathrm{8}×\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} ,\mathrm{8}×\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} ,\mathrm{12},\mathrm{12}×\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} ,\mathrm{12}×\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} ,\mathrm{18},\mathrm{18}×\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} ,\mathrm{18}×\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} ,\mathrm{27}…. \\ $$

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