Consider the sequence provided below 3^x, 3^3x, 3^5x, 3^7x,... Is the Given sequence a GP? give a reason, if it is a GP, find the 21st term
Yes, the given sequence is a geometric progression (GP).
In a geometric progression, each term is obtained by multiplying the previous term by a common ratio. Let's find the common ratio (r) in this sequence:
To find the common ratio between consecutive terms, we divide a term by its previous term:
3^3x / 3^x = (3^3x) / (3^x) = 3^(3x - x) = 3^(2x)
So, the common ratio (r) in this GP is 3^(2x).
Now, to find the 21st term, we can use the formula for the nth term of a GP:
tn = ar^(n-1)
where tn is the nth term, a is the first term, r is the common ratio, and n is the position of the term.
In this sequence, the first term a is 3^x, the common ratio r is 3^(2x), and we want to find the 21st term, so n = 21.
t21 = (3^x) * (3^(2x))^(21-1)
= (3^x) * 3^(40x)
= 3^(x + 40x)
= 3^(41x)
Therefore, the 21st term of the given sequence is 3^(41x).
To determine if the given sequence is a geometric progression (GP), we need to check if the ratio of any two consecutive terms is constant.
Let's calculate the ratio of any two consecutive terms:
Ratio between the 2nd and 1st terms: (3^3x) / (3^x) = 3^(3x - x) = 3^(2x)
Ratio between the 3rd and 2nd terms: (3^5x) / (3^3x) = 3^(5x - 3x) = 3^(2x)
Ratio between the 4th and 3rd terms: (3^7x) / (3^5x) = 3^(7x - 5x) = 3^(2x)
From the calculations, we can see that the ratio between any two consecutive terms is 3^(2x). Since this ratio is constant for all pairs of terms, we can conclude that the given sequence is indeed a geometric progression.
To find the 21st term, we can use the formula for the nth term of a geometric progression:
Tn = a * r^(n-1)
In this case, the 1st term (a) is 3^x and the common ratio (r) is 3^(2x). Plugging these values into the formula, we get:
T21 = (3^x) * (3^(2x))^(21-1)
= (3^x) * (3^(2x))^20
= (3^x) * 3^(40x)
= 3^(x + 40x)
= 3^(41x)
Therefore, the 21st term of the given sequence is 3^(41x).