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Question: If \( x > 1,y > 1,z > 1 \) in G.P then, \( \dfrac{1}{{1 + lnx}},\dfrac{1}{{1 + lny}},\dfrac{1}{{1 + ...

If x>1,y>1,z>1x > 1,y > 1,z > 1 in G.P then, 11+lnx,11+lny,11+lnz\dfrac{1}{{1 + lnx}},\dfrac{1}{{1 + lny}},\dfrac{1}{{1 + lnz}} are in
A) A.P
B) H.P
C) G.P
D) none of above

Explanation

Solution

Hint : Greater than and less than symbols can be used to compare numbers and expressions. The greater than symbol looks like >> . It is a series whose reciprocals are in arithmetic progression. An arithmetic progression (AP) or arithmetic progression sequence of numbers such that the difference between the consecutive terms is constant.

Complete step-by-step answer :
Let, x,y,zx,y,z are in G.P
Let, the common ratio of the G.P be rr
Then, y=xry = xr and z=xr2z = x{r^2}
So, let the see what is achieved by putting price in x,y,zx,y,z
lny=lnx+lnr\Rightarrow lny = lnx + lnr and lnz=lnx+2lnrlnz = lnx +2lnr
Let, A=1+lnxA = 1 + lnx and D=lnrD = lnr
Then 11+lnx=1A\dfrac{1}{{1 + lnx}} = \dfrac{1}{A}
Therefore,
11+lny=11+lnx+lnr=1A+D\dfrac{1}{{1 + lny}} = \dfrac{1}{{1 + lnx + lnr}} = \dfrac{1}{{A + D}}
Now, same as it is we know that
11+lnz=11+lnz+2lnr=1A+2D\dfrac{1}{{1 + lnz}} = \dfrac{1}{{1 + lnz + 2lnr}} = \dfrac{1}{{A + 2D}}
Therefore, 11+lnx,11+lny,11+lnz\dfrac{1}{{1 + lnx}},\dfrac{1}{{1 + lny}},\dfrac{1}{{1 + lnz}} are in H.P
So, the correct answer is “Option B”.

Note : If you are having trouble identifying the nth term and n. then it should be clear to you that in place of the nth term , you must put the value of nth term and in the place of n you have to put the number of that term .When you are not able to decide which formula will be applicable Know that there are only three things that determine an AP, first term, common difference and number of terms knowing this you can find an AP , then depending on the data given apply the appropriate formula and find the unknowns also be careful with calculations. The general form of a G.P is a,ar,ar2,ar3a,ar,a{r^2},a{r^3} and so on. The nth{n^{th}} term of G.P series is Tn=arn1Tn = a{r^{n - 1}} ,
Where, a=a = first term and r=r = common ratio .