If the fourth term of GP is 9 and the sixth term is 81.
Find:
1) common ratio.
2) first term
ar^3 = 9
ar^5 = 81
divide them:
r^2 = 9
r = ± 3
take the two cases for r, sub into one of the equations to solve for a
(two sets of answers)
I don't see the value of first term of geometric progression
I don’t know
1/3
To find the common ratio (r) of a geometric progression (GP), we can use the formula:
\[ \text{term}_n = a \times r^{(n-1)} \]
where "term_n" represents the nth term of the GP, "a" is the first term, and "r" is the common ratio.
1) To find the common ratio:
Given that the fourth term is 9, we can substitute n = 4 into the formula:
\[ 9 = a \times r^{(4-1)} \]
\[ 9 = a \times r^3 \]
Similarly, for the sixth term, we can substitute n = 6 into the formula:
\[ 81 = a \times r^{(6-1)} \]
\[ 81 = a \times r^5 \]
We now have a system of equations:
\[ \begin{cases} 9 = a \times r^3 \\ 81 = a \times r^5 \end{cases} \]
To solve this system, we can divide the second equation by the first equation:
\[ \frac{81}{9} = \frac{a \times r^5}{a \times r^3} \]
Simplifying:
\[ 9 = r^2 \]
Taking the square root of both sides:
\[ r = \pm 3 \]
Since the common ratio of a GP cannot be negative, we have:
\[ r = 3 \]
2) To find the first term (a):
Now that we know the common ratio (r = 3), we can substitute either of the given equations to find the value of the first term (a).
Let's use the equation for the fourth term (n = 4):
\[ 9 = a \times 3^3 \]
\[ 9 = a \times 27 \]
Solving for "a":
\[ a = \frac{9}{27} \]
\[ a = \frac{1}{3} \]
Therefore, the common ratio (r) is 3 and the first term (a) is \( \frac{1}{3} \).