The sum of 4th and 6th terms of a geometric series is 80. If the product of the 3rd and 5th term is 256 determine first term and common ratio.
If oobleck had not noticed that particular sum of the two powers of 2 ...
ar^3 + ar^5 = 80 ---> ar^3( 1 + r^2) = 80
ar^2 * ar^4 = 256 ---> a^2 r^6 = 256 ---- > a r^3 = 16
sub that into the first:
16(1+ r^2) = 80
1+r^2 = 8
r^2 = 4
r = ± 2
If r = 2, a(8) = 16, a = 2
if r + -2, a(-8) = 16, a = -2
for r = 2 and a = 2, the sequence is : 2, 4, 8, 16, 32, 64, ....
for r = -2 and a = -2, the sequence is: -2, 4, -8, 16, -32, 64, ...
Not understandable clearly
ar^3 + ar^5 = 80
ar^2 * ar^4 = 256
Note that
80 = 16+64 = 2*(2^3+2^5)
256 = 2^8 = 2^2 * 2^6
so it looks like
a = 2, r=2
and the GP is 2,4,8,16,32,64,...
let t equal the 3rd term
t r + t r^3 = t r (1 + r^2) = 80
t * t r^2 = (t r)^2 = 256 ... t r = 16
substituting ... 16 (1 + r^2) = 80 ... 1 + r^2 = 5 ... r = 2 ... t = 8
t = (1st) * r^2 ... 1st = 8 / (2^2) = 2
i didn`t get the point clearly
To solve this problem, we need to use the formula for the n-th term of a geometric series, which is given by:
\[a_n = a_1 \cdot r^{(n-1)}\]
where \(a_n\) is the n-th term, \(a_1\) is the first term, \(r\) is the common ratio, and \(n\) is the term number.
Let's break down the problem step by step:
1. The sum of the 4th and 6th terms of the geometric series is 80. We can write this as an equation:
\[a_4 + a_6 = 80\]
Substituting the formula for the n-th term, we get:
\[a_1 \cdot r^3 + a_1 \cdot r^5 = 80\]
Factoring out \(a_1\), we have:
\[a_1 \cdot (r^3 + r^5) = 80\] ---(equation 1)
2. The product of the 3rd and 5th terms is 256. We can write this as an equation:
\[a_3 \cdot a_5 = 256\]
Using the formula for the n-th term, we substitute and get:
\[(a_1 \cdot r^2) \cdot (a_1 \cdot r^4) = 256\]
Simplifying, we have:
\[a_1^2 \cdot r^6 = 256\] ---(equation 2)
Now, we have two equations (equation 1 and equation 2) with two unknowns ( \(a_1\) and \(r\) ). We can solve this system of equations simultaneously.
To eliminate \(a_1\), we can divide equation 1 by equation 2:
\[\frac{a_1 \cdot (r^3 + r^5)}{a_1^2 \cdot r^6} = \frac{80}{256}\]
Simplifying, we get:
\[\frac{(r^3 + r^5)}{(a_1 \cdot r^5)} = \frac{5}{16}\]
Since we know that \(a_1 \neq 0\) (geometric series cannot have a first term of 0) we can cancel out the \(a_1\) terms and simplify further:
\[r^3 + r^5 = \frac{5}{16} \cdot r^5\]
Now, we have a cubic equation in terms of \(r\):
\[r^3 + r^5 - \frac{5}{16} \cdot r^5 = 0\]
This equation can be solved using numerical methods or algebraic techniques. Once we find the value of \(r\), we can substitute it back into equation 1 to solve for \(a_1\).