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shaik0124Best ResponseYou've already chosen the best response.2
If a, b, and c are in GP, then b = ak and c = ak^2 for some constant k. This gives: log₂(b) = log₂(ak) = log₂(a) + log₂(k) log₂(c) = log₂(ak^2) = log₂(a) + 2log₂(k). This shows that log₂(a), log₂(b), and log₂(c) form a AP with a first term of log₂(a) and a common difference of log₂(k). Therefore, the answer is a) in AP. I hope this helps!
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
But My book says it is H.P
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
we have ac = b^2 and we need to find about 1/loga , 1/log b , 1/log c where all bases are converted to x now clearly, log a + log c = log(ac) = 2logb thus log a,log b and log c are in AP so 1/log a , 1/log b and 1/log c are in HP
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
SINCE they are in G.P b^2 = ac or, logac/logb = 2 or, loga + log c = 2 logb now, logb= loga+logc /2
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
the 3rd last line can also be understood if you take log to base x on both sides of ac= b^2
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
Thus they are in A.P
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
exactly!! as @sauravshakya said/did..
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
They will be in Harmonic Progression..
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
Yup....they should be in HP ... But hw
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
ohh,,didnt i explain well @Yahoo! ?
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
Let me write.. It will take time.
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
@shubhamsrg finally u agreed that it is Ap
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
so 1/log a , 1/log b and 1/log c are in HP
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
lol..noo..i didnt.. i meant you can derive loga ,log b and log c are in AP but @sauravshakya 's method..
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
which ultimately means the ans is HP as you are looking for 1/log a,1/log b and 1/log c
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
Let: \[\large \log_a(x) = k \implies a^k = x \implies a = x^{\frac{1}{k}}\] Similarly: Let others: \[\large b = x^{\frac{1}{m}}\] \[\large c = x^{\frac{1}{n}}\]
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
Now given: \[b^2 = ac\]
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
\[\huge x^{\frac{2}{m}} = x^{\frac{1}{k}} \times x^{\frac{1}{n}}\]
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
\[\implies \frac{1}{k} + \frac{1}{n} = \frac{2}{m}\] This implies k, m and n are in HP..
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
How substitute back for k, m and n back..
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
\[k = \log_a(x), m = \log_b(x), n = \log_c(x)\]
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
Can we subs the value for a b c such as 2 , 4 , 8 and assume value for x as 8 and solve the log??
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
This you can do if you want to verify..
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
i did by this and got answer as it is AP
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
Let me check then..
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
dw:1344689783197:dw right?
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
Yup.. Base changing rule
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
They are in HP @Yahoo!
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
Let a = 2, b= 4 and c = 8 Let x = 64
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
dw:1344689862033:dwsimilarly....
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
\[\log_2(64) = 6\] \[\log_4(64) = 3\] \[\log_8(64) = 2\] \[\huge 6, 3, 2 \; \; are \; \; In \; \; HP..\]
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
\[\frac{2}{3} = \frac{1}{6} + \frac{1}{2}\]
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
There is no condition you can take x = 8.. But the fact is that by taking 8 it will not become AP..
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
log2 (8( = 4 log4(8) = 2 log 8 (8) = 1
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
\[\huge \log_4(8) \ne 2\]
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
Now,SINCE they are in G.P b^2 = ac or, logac/logb = 2 or, loga + log c = 2 logb now, logb= loga+logc /2 Thus,1/log a , 1/log b and 1/log c are in HP So, logx/loga , logx/logb and logx/logc are in HP thus, loga (x) , logb (x) , logc (x) are in HP
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
\[\huge \log_4(8) =\frac{3}{2}\]
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
Use the base change property that @sauravshakya gave you..
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
oh..ok.... i really need to study logarithms
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
@waterineyes @sauravshakya @shubhamsrg thxxxxxx
 one year ago
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