The fifth term of an exponential sequence (g.p)is greater than the first term by 13 and the half and the first term is greater than the third term by 9.find the common ratio and the first term
To solve this problem, we need to use the formula for the nth term of an exponential (geometric progression) sequence which is given by:
An = A * r^(n-1)
where An represents the nth term, A represents the first term, r represents the common ratio, and n represents the position of the term.
We are given two pieces of information:
1. The fifth term is greater than the first term by 13, so A + 13 = A * r^4
2. The first term is greater than the third term by 9, so A - 9 = A * r^2
We can use these two equations to find the value of the common ratio (r) and the first term (A).
Step 1: Simplify the equations
We can rewrite the equations as follows:
(A * r^4) - A = 13
(A * r^2) - A = 9
Step 2: Solve for the common ratio (r)
Subtracting A from both sides of the equation gives:
A * r^4 - A = 13
A * r^2 - A = 9
Divide the equations to eliminate A:
(A * r^4 - A) / (A * r^2 - A) = 13 / 9
Simplify the equation:
r^2 + 1 = (13/9)
Step 3: Solve for the first term (A)
Substitute the value of r back into either of the original equations to find A.
Using A * r^2 - A = 9, substitute r^2 = (13/9) - 1:
A * ((13/9) - 1) - A = 9
Simplify the equation:
(4/9) * A = 9
Multiply both sides by 9/4 to solve for A:
A = (9 * 9) / 4
Simplify:
A = 81/4
Therefore, the common ratio (r) is the square root of (13/9 - 1) and the first term (A) is 81/4.
Hmmm. I suspect a typo.
If the 5th term is greater than the 1st, then |r| > 1
But if the 1st term is greater than the 3rd, then |r| < 1