The sum of the last three terms of a G.P having n terms is 1024 times the sum of the first three terms of the progression. If the third term is 5, find the last term.
the third term is 5 ----> a r^2=5
a = 5/r^2
1024(a + ar + ar^2) = ar^(n-3) + ar^(n-2) + ar^(n-1)
divide by a
1024(1 + r + r^2) = r^(n-3) + r^(n-2) + r^(n-1)
1024(1 + r + r^2) = r^(n-3)(1 + r + r^2)
1024 = r^(n-3)
we know: 2^10 = 1024 OR 4^5 = 1024
case1
let r = 2 , then
n-3 = 10
n=13
a = 5/2^2 = 5/4
sequence is 5/4, 5/2, 5, 10, 20, 40, 80, 160, 320, 640, 1280, 2560, 5120
The last term is ar^12 = 5120
check: 1024(5/4 + 5/2 + 5) = 8960
1280+2560+5120 = 8960
case2 , r = 4, n = 8
terms are 5/16, 5/4, 5, 20, 80, 320, 1280, 5120
1024(5/16 + 5/4 + 5) = 6720
320+1280+5120 = 6720
both cases work,
anyway, the last term in each case is 5120
S(n)-S(n-3) = 1024S(3)
(a(r^n-1)-a(r^(n-3)-1)/(r-1) = 1024(r^3-1)/(r-1)
(r^n-1)-(r^(n-3)-1) = 1024(r^3-1)
r^(n-3)(r^3-1) = 1024(r^3-1)
r^(n-3) = 1024 = 2^10
So, let's say r=2 and n=13
Then the sum of the last 3 terms is
2^11+2^12+2^13 = 2^10(2+4+8)
and the sum of the 1st 3 terms is 2+4+8 = 14
But we want the 3rd term to be 5, so a = 5/14
and the 13th term is 5/14 * 2^13 = 5*4096/7
I botched it at the end. I'm sure you can see where.
Good work, Reiny, as usual!
To find the last term of the geometric progression (G.P.), we need to use the given information and apply the formula for the sum of a geometric series.
Let's denote the first term of the G.P. as 'a', and the common ratio as 'r'. We are given that the third term is 5, so we can write the third term as:
a * r^2 = 5 ----(1)
We are also given that the sum of the last three terms is 1024 times the sum of the first three terms. For a G.P., the sum of the first three terms is:
S3 = a + ar + ar^2 ----(2)
The sum of the last three terms is:
S_n = ar^(n-2) + ar^(n-1) + ar^n
By dividing the sum of the last three terms by the sum of the first three terms, we get:
S_n / S3 = (ar^(n-2) + ar^(n-1) + ar^n) / (a + ar + ar^2)
Since we are given that this ratio is equal to 1024, we can write:
(ar^(n-2) + ar^(n-1) + ar^n) / (a + ar + ar^2) = 1024 ----(3)
We have two equations (1) and (3) with two variables (a and r). Using these equations, we can solve for the values of a and r.
First, let's solve equation (1) for a in terms of r:
a = 5 / r^2
Substituting this value of a into equation (3), we get:
[(5r^(n-2)) + (5r^(n-1)) + (5r^n)] / [5 + 5r + 5r^2] = 1024
Next, let's simplify the equation:
(5r^(n-2) + 5r^(n-1) + 5r^n) = 1024(5 + 5r + 5r^2)
5(r^(n-2) + r^(n-1) + r^n) = 1024(5 + 5r + 5r^2)
Dividing both sides by 5:
r^(n-2) + r^(n-1) + r^n = 1024(r^2 + r + 1)
Now let's consider the specific case where the third term is 5, which we know from equation (1):
r^2 = 5 / a = 5 / (5 / r^2) = r^4
Substituting r^2 = r^4 into the equation:
r^(n-2) + r^(n-1) + r^n = 1024(r^4 + r^2 + 1)
Now let's simplify the equation further. Since we have r^(n-2) + r^(n-1) + r^n = (r^2)^(n-2) + (r^2)^(n-1) + (r^2)^n, we can substitute r^2 for r^4:
(r^2)^(n-2) + (r^2)^(n-1) + (r^2)^n = 1024(r^4 + r^2 + 1)
Simplifying the exponents:
r^(2n-4) + r^(2n-2) + r^(2n) = 1024(r^4 + r^2 + 1)
Now we have an equation with only 'r' terms. However, solving this equation to find the value of 'r' for a given 'n' would involve a lengthy process of finding roots and manipulating exponents. Unfortunately, the equation does not simplify easily, and there is no straightforward way to solve it algebraically.
Therefore, to find the last term of the G.P., we need additional information or conditions, or we need to be given a specific value for 'n'. Without this additional information, we cannot determine the last term of the G.P.