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Question-123869




Question Number 123869 by benjo_mathlover last updated on 28/Nov/20
Answered by liberty last updated on 29/Nov/20
let :a ,ar,ar^2 ,ar^3  the four consecutive terms in GP  given the condition Σ_(i=1) ^4 u_i  = a+ar+ar^2 +ar^3 =80  a(1+r+r^2 (1+r))=80 ; a(1+r)(1+r^2 )=80  and ((ar+ar^3 )/2)=30 ; ar(1+r^2 )= 60   ⇒ ((a(1+r)(1+r^2 ))/(ar(1+r^2 ))) = ((80)/(60))=(4/3) , we get 3+3r=4r  r = 3 then a = ((60)/(3×10)) = 2.   thus the terms is 2, 6, 18, 54.
$${let}\::{a}\:,{ar},{ar}^{\mathrm{2}} ,{ar}^{\mathrm{3}} \:{the}\:{four}\:{consecutive}\:{terms}\:{in}\:{GP} \\ $$$${given}\:{the}\:{condition}\:\underset{{i}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{u}_{{i}} \:=\:{a}+{ar}+{ar}^{\mathrm{2}} +{ar}^{\mathrm{3}} =\mathrm{80} \\ $$$${a}\left(\mathrm{1}+{r}+{r}^{\mathrm{2}} \left(\mathrm{1}+{r}\right)\right)=\mathrm{80}\:;\:{a}\left(\mathrm{1}+{r}\right)\left(\mathrm{1}+{r}^{\mathrm{2}} \right)=\mathrm{80} \\ $$$${and}\:\frac{{ar}+{ar}^{\mathrm{3}} }{\mathrm{2}}=\mathrm{30}\:;\:{ar}\left(\mathrm{1}+{r}^{\mathrm{2}} \right)=\:\mathrm{60}\: \\ $$$$\Rightarrow\:\frac{{a}\left(\mathrm{1}+{r}\right)\left(\mathrm{1}+{r}^{\mathrm{2}} \right)}{{ar}\left(\mathrm{1}+{r}^{\mathrm{2}} \right)}\:=\:\frac{\mathrm{80}}{\mathrm{60}}=\frac{\mathrm{4}}{\mathrm{3}}\:,\:{we}\:{get}\:\mathrm{3}+\mathrm{3}{r}=\mathrm{4}{r} \\ $$$${r}\:=\:\mathrm{3}\:{then}\:{a}\:=\:\frac{\mathrm{60}}{\mathrm{3}×\mathrm{10}}\:=\:\mathrm{2}.\: \\ $$$${thus}\:{the}\:{terms}\:{is}\:\mathrm{2},\:\mathrm{6},\:\mathrm{18},\:\mathrm{54}.\: \\ $$

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