Elitmus
Exam
Logical Reasoning
Number Series
G1,G2,G3,G4 are in G.P, where G=G1 and R is the common ration and log4(G1)+log4 G2 +log4 G3 +.... = 2500, then find the possible pair of (G,R).
Read Solution (Total 7)
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- G1,G2,G3,G4 are in G.P
G1=G, G2=G*R , G3=G*R^2 , G4=G*R^3
log4(G1)+ log4 G2 +log4 G3 +log4 G4 = 2500
=> log4(G1*G2*G3*G4)=2500
=> G1*G2*G3*G4 = 4^2500
=> G*GR*GR^2*GR^3= 4^2500
=> G^4*R^6= 4^2500
=> G^2*R^3= 2^2500
(G,R)=[ 2^1250/R^(3/2) , R]
IF R=1, G^2=2^2500 => G=2^1250
IF R=2^2, G^2=2^2494 => G=2^1247
IF R=2^4, G^2=2^2488 => G=2^1244
and so on........ - 10 years agoHelpfull: Yes(13) No(6)
- how is it possible
(G,R)=[ 2^1250/R^(3/2) , R]
IF R=1, G^2=2^2500 => G=2^1250
IF R=2^2, G^2=2^2494 => G=2^1247
IF R=2^4, G^2=2^2488 => G=2^1244
and so on........ - 10 years agoHelpfull: Yes(2) No(0)
- Rakesh's solution is right till the expression (G,R)=[ 2^1250/R^(3/2) , R]. He wend wrong on his assumptions thereafter. We could get the answer by putting all values in the options for R and finding G.
- 10 years agoHelpfull: Yes(0) No(1)
- provide a valid solution
- 10 years agoHelpfull: Yes(0) No(1)
- are there only 4 terms in series or is it upto n Terms if just 4 terms then G=4^622
R=4^2 - 10 years agoHelpfull: Yes(0) No(0)
- R =G=4^250
- 10 years agoHelpfull: Yes(0) No(1)
- Given G1,G2,G3,G4 are in G.P.
Let G1=G,G2=GR,G3=G(R)^2,G4=G(R)^3
Log(G1)base4+.........................+Log(G4)base4=2500
Log(G1G2G3G4)=2500 putting values of G1,G2,G3,G4
(G^4) (R^6)=42500
(G,R)=[(2^1250)/(R^(2/3)) ,R] - 7 years agoHelpfull: Yes(0) No(0)
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