1) The sum of last three terms of GP having n terms is 1024 time the sume of first 3 terms of GP. If 3rd term is 5. Find the last term.
2) The sum of the first eight terms of a GP is five times the sum of the first four terms. Find the common ratio.
(1) ar^(n-1) + ar^(n-2) + ar^(n-3) = 1024 (a + ar + ar^2)
ar^(n-3) (1 + r + r^2) = 1024a (1 + r + r^2)
r^(n-3) = 1024
It is known that the third term is 5, so: ar^2 = 5
The last term will be:
ar^(n-1) = ar^2. r^(n-3) = 5 x 1024 = ? (You can finish the rest of it, right?)
(2) S8 = 5 S4
a(r^8 - 1)/(r-1) = 5a(r^4 - 1)/(r - 1)
(r^4 - 1)(r^4 + 1) = 5(r^4 - 1)
r^4 + 1 = 5
r^4 = 4
r = Âħsqrt(2)
To solve these geometric progression (GP) problems, we need to use the formulas for the nth term and the sum of the first n terms of a GP.
1) Let's find the common ratio, r, of the given GP:
The third term is given as 5, so we can write the third term as a2 = a1 * r, where a1 is the first term.
We can also write the fourth term, a3, as a3 = a1 * r^2, and the fifth term, a4 = a1 * r^3.
Now, we can write the sum of the first three terms, S3, as S3 = a1 + a2 + a3 = a1 + a1 * r + a1 * r^2.
Similarly, the sum of the last three terms, S_last3, is S_last3 = a(n-2) + a(n-1) + a_n = a1 * r^(n-3) + a1 * r^(n-2) + a1 * r^(n-1).
Since we're given that S_last3 = 1024 * S3, we can write the equation:
a1 * r^(n-3) + a1 * r^(n-2) + a1 * r^(n-1) = 1024 * (a1 + a1 * r + a1 * r^2).
We're also given that a3 = 5, so we can substitute this to get: a1 * r^2 = 5.
Now, let's use this information to solve for the last term, a_n:
To find the last term, a_n, we need to find n first. Since we know a3 = 5, we can solve the equation a1 * r^2 = 5 for a1:
a1 = 5 / r^2.
Substituting this into the equation for S3, we get:
S3 = a1 + a1 * r + a1 * r^2
S3 = 5 / r^2 + (5 / r^2) * r + (5 / r^2) * r^2
S3 = 5 / r^2 + 5 / r + 5
S3 = (5 + 5r + 5r^2) / r^2.
Now we know that S_last3 = 1024 * S3, so we can write:
(a1 * r^(n-3) + a1 * r^(n-2) + a1 * r^(n-1)) = 1024 * (a1 + a1 * r + a1 * r^2)
(a1 * r^3 + a1 * r^4 + a1 * r^5) = 1024 * (a1 + a1 * r + a1 * r^2)
(5 * r^3 + 5 * r^4 + 5 * r^5) = 1024 * ((5 + 5r + 5r^2) / r^2).
Now we can simplify this equation:
5 * r^3 + 5 * r^4 + 5 * r^5 = 1024 * (5 + 5r + 5r^2) / r^2
Multiply both sides by r^2 to get rid of the denominator:
5 * r^5 + 5 * r^6 + 5 * r^7 = 1024 * (5 + 5r + 5r^2).
This equation will give us r in terms of its powers, but we can use numerical methods or an algebraic calculator to solve this equation. Solving this equation gives us the value of r, which we can then use to find the last term, a_n, using the formula a_n = a1 * r^(n-1).
2) Similarly, for the second question, we're given that the sum of the first eight terms, S8, is five times the sum of the first four terms, S4. We can write this as:
S8 = 5 * S4.
Using the formula for the sum of the first n terms of a GP, we have:
S8 = a1 * (r^8 - 1) / (r - 1)
S4 = a1 * (r^4 - 1) / (r - 1).
Substituting these equations into S8 = 5 * S4, we get:
a1 * (r^8 - 1) / (r - 1) = 5 * (a1 * (r^4 - 1) / (r - 1)).
We can now simplify this equation:
(r^8 - 1) / (r - 1) = 5 * (r^4 - 1) / (r - 1).
Canceling out the denominator, we have:
r^8 - 1 = 5 * (r^4 - 1).
Now, using algebraic techniques or numerical methods, we can solve this equation to find the value of r, which will be the common ratio of the given GP.
After finding the common ratio, we can use this value to find the other terms in the GP by using the formula a_n = a1 * r^(n-1).