Find the number of ways in which 3 distinct numbers can be selected from the set $\{3^1,3^2,\dots,3^{100},3^{101}\}$ so that they form gp.. Find the number of ways in which $3$ distinct numbers can be selected from the set $\{3^1,3^2,\dots,3^{100},3^{101}\}$ so that they form gp..
My attempt.
I tried to take numbers like $3^1,3^3,3^5$ (whose powers have a common difference of $2$ then $3$ then $4$, and so on but that's lengthy..
2nd Attempt
we know for numbers to be in GP (geometrical progression)..
$b^2=ac$
we can write $b$ as some $3^k$, $a$ as $3^l$, and $c$ as $3^m$
We get the $2k=l+m$.
Now we have to find the triplets which are in ap(arithmetical progression) from the set $\{1,2,3,\dots,100,101\}$. How to do this.??
 A: Let's separate this and look this way. How many geometric series of $3$ numbers can be selected so the common ratio will be $3$?
Every number can become first term of the series, except the last two, which means that there are $99$ ways to select geometric series of $3$ numbers.
Now let's the ratio be $3^2$. We can easily see that every number except the last four can be first term of the series.
Can you spot the pattern.
Here's the general formula:
$$\text{Number of combinations} = \sum_{n=1}^{\left\lfloor\frac{k}{2}\right\rfloor} k-2n$$
Actually if $k$ is even then the number of combinations is the sum of all even numbers smaller than $k$ and if $k$ is odd number of combinations is the sum of all odd numbers smaller than $k$.
If $k$ is even then:
$$\text{Number of combinations} = \left(\frac{k}{2} - 1\right)\left(\frac{k}{2}\right)$$
If $k$ is odd then:
$$\text{Number of combinations} = \left(\frac{k-1}{2}\right)^2$$
A: Lets pick the end terms, in such a way that the middle term is bound to be there as well.
If you're going to take $3^m$  and $3^n$ as your first and third terms, then the middle term, $T_2$ will be $$T_2=3^{\frac{m+n}{2}}$$For  $T_2$ to be in the set, $(m+n)$ must be even; $m$ and $n$ must have the same parity.
There are $50$ set members with even exponents on the $3$. Two can be selected in $\binom{50}{2}$ ways.
There are $51$ set members with odd exponents on the $3$. Two can be selected in $\binom{51}{2}$ ways.
A: As you have concluded, the number should be equal to the number of triplets (with distinct elements) which are in arithmetic progression in the set $\{1,2,3,...,101\}$.
Consider all triplets with common difference, $d=1$. We are counting the triplets $(1,2,3),\,(2,3,4)...(99,100,101)$.  There are $101-2=99$ such triplets.
For $d=2$, we are counting $(1,3,5),(2,4,6)$ etc. and we have $101-4=97$ such triplets. Similarly, for $d=3$, the number is $101-6$ and so on. 
The biggest triplet is $(1,51,101)$, which is for $d=50$.
So, we are looking for $\sum_{d=1}^{d=50}101-2d$.
A: Hint: Find any two numbers with an even difference. How many ways can this be done? What will the other number in the triplet be (this relates to the even difference part)?
