Question

give the solution of the question

Answer 1

3, 6, 12 are in GP { because you can see the common ratio is 2 (constant)}

now, take log base 2,
log{3 base 2} , { log{6 base 2}, log{12 base 2} are in AP .
how? Let see
log{3 base 2} = first term

log{6 base 2} = log{ 3×2 base 2}
= log{3 base 2} + log{ 2 base 2}
= log{3 base 2} + 1 = second term

log{12 base 2} = log{3×2² base 2}
= log{3 base 2} + 2log{2 base 2}
= log{3 base 2} +2 = 3rd term

we see that,
log{3 base 2}, log{3base 2} +1 , log{3 base 2} + 2 ,are in AP

hence,
log{3 base 2}, log{6 base 2} , log{12 base 2} are in AP
so,
1/log{3 base 2} , 1/log{6 base 2} , 1/log{12 base 2} are in HP

we know, a/c to properties of logrithmn ,
log{a base b} = 1/log{b base a} use this here,

1/{1/log(2 base 3) }, 1/{1/log(2 base 6)}, 1/log(2 base 12)} are in HP
so,
log(2 base 3), log(2 base 6) , log(2 base 12) are HP.
hence,
1/log(2 base 3) , 1/log(2 base 6) , 1/log(2 base 12) are in AP

hence proved //

[ note :- inverse of AP is in HP and vice-versa ]

Answer 2

Step-by-step explanation:

1/log2 base 3=log3 base 2

1/log2 base 6=log 6 base 2

1/log 2 base 12=lig12 base 2

Then,in ap all common differences are same

So log 6 base 2- log 3 base 2=log 12 base 2 - log 6 base 2

Bases are same so, log (6/3) base 2=log (12/6) base 2

Log 2 base 2 = log 2 base 2

1=1

Common differences are same

So they are in A. P

hence proved