IIT JEE , AIEEE Forum - Mathematics , Algebra

achinbansal

solve this

If log a, log b ,log c are in AP and also log a-log 2b ,log2b-log3c , log3c-loga are in AP then

abc are in ap

a 2b 3c are in ap

abc are the sides of the triangle

none

please give me the comp soln

i will be grateful

prash_shan_jpr

is d ans is sides of a triangle?? tell me i will post d solution..

LAMPARD

yes,they are sides of triangle.
From 1st condition,
logb= (loga + logc)/2
Thus, b²=ac........(1)
From 2nd condition,
log2b-log3c = (log3c - log2b)/2
Thus,
2b=3c or c=2b/3
Subst. c in (1) to get c=4a/9.
Thus,we can see that they the first two options are not satisfied.
Clearly,its the third option,i.e,they are sides of triangle.

spideyunlimited

log a, log b ,log c are in AP
loga + logc = 2logb
logac = logb^2
ac = b^2 .....................so a,b,c are in GP.

log a-log 2b ,log2b-log3c , log3c-loga are in AP
loga - log2b + log3c - loga = 2 (log2b - log3c)
log(3c/2b) = 2 (log (2b/3c))
3c/2b = (2b/3c)^2
THUS, 3c/2b = 1
3c = 2b
b = 3c/2

ac= b^2 = 9c^2/4
a = 9c/4
b = 3c/2
c = c

since this is a pythagorean triplet (sum of any 2 is greater than 3rd)
so they are the sides of a triangle.

me_living4u

a,b,c > 0

log a + log c = 2log b

=> ac = b^2............................ 1

and

as log a+ log b= log ba...and

log a - log b = log a/b

log a/2b + log 3c/a = log 2b/3c

=> log 3c/2b = log 2b/3c

=> 2b= 3c as all r postive............. 2

solving frm 1 and 2

4a= 6b= 9c=k

so a=k/4= 36m/4=9m.... b=k/6= 6m.... c = 4m...whre let k= 36m( lcm of 4,6,9)

now sum of any two > third

lik a+b>c or

b+c> a

c+a> b

so these are sides of a triangle

getting unusual ans.

hope its rit

Cheers!!!

prash_shan_jpr

Re:solve this

achinbansal

yes the ans is correct

even after studying so much hoow cant i solve such easy prob

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