find the sequence whose n th term is 1/3n-2. is this sequence an H.P.?
since the reciprocals are in AP, yes, it is an HP.
Please find the sequence whose n th term is 1/3n-2 is this sequence an H.P.
Oh, I see, you're looking for the sequence whose n-th term is 1/3n-2. Well, let's analyze it, shall we?
The n-th term of the sequence is given by 1/3n-2. If we simplify it a bit, we get 1/(3n) - 2.
Now, an arithmetic progression (A.P.) is a sequence in which the difference between consecutive terms is constant. If we check the difference between consecutive terms of your sequence, we'll find it is not constant, as it involves division and subtraction.
So, to answer your question, no, this sequence is not an H.P. It's more like a sneaky mixture of fractions and algebraic expressions. Just like trying to make a salad with jellybeans and rubber duckies, it doesn't quite fit the traditional definitions. Keep on searching for that elusive H.P. sequence, my friend!
To determine if the given sequence is a Harmonic Progression (H.P.), we need to check if the reciprocals (1/nth term) form an arithmetic sequence.
The nth term of the given sequence is: 1/(3n - 2).
To find the reciprocals, we take the reciprocal of each term:
Reciprocal of 1/(3n - 2) = (3n - 2)/1 = 3n - 2.
Now, let's check if the reciprocals form an arithmetic sequence:
To do that, we calculate the difference between consecutive terms. Let's calculate the difference between the (n+1)th term and nth term:
(3(n+1) - 2) - (3n - 2)
= 3n + 3 - 2 - 3n + 2
= 1.
Since the difference between consecutive terms is a constant (1), we can conclude that the reciprocals of the given sequence form an arithmetic sequence. Hence, the given sequence is a Harmonic Progression (H.P.).