A geometric progression (G.P) has common ratio 2. Find the value of 𝑛 for which the sum of
2𝑛 terms is 33 times the sum of the first 𝑛 terms.
Let the first term of the geometric progression be a. Then the sum of the first n terms is:
S_n = a(1 - r^n)/(1 - r)
where r = 2 is the common ratio. Similarly, the sum of the first 2n terms is:
S_2n = a(1 - r^(2n))/(1 - r)
We are given that:
S_2n = 33S_n
Substituting the formulas for S_n and S_2n, we get:
a(1 - r^(2n))/(1 - r) = 33a(1 - r^n)/(1 - r)
Simplifying, we get:
1 - 2^(2n) = 33(1 - 2^n)
Expanding and simplifying again, we get:
2^n(32 - 2^n) = 32
Letting x = 2^n, we can rewrite this equation as:
x(32 - x) = 32
Simplifying, we get:
x^2 - 32x + 32 = 0
Using the quadratic formula to solve for x, we get:
x = (32 ± sqrt(32^2 - 4*32))/2
x = 16 ± 8sqrt(3)
Since x = 2^n, we must have:
2^n = 16 ± 8sqrt(3)
Taking the logarithm base 2 of both sides, we get:
n = log2(16 ± 8sqrt(3))
Using a calculator, we find:
n ≈ 4.09 or n ≈ 6.91
Since n must be an integer, the only solution is n = 6.
Let's say the first term of the geometric progression is 'a' and the common ratio is 'r'.
The sum of the first n terms of a geometric progression is given by the formula:
S_n = a * (1 - r^n) / (1 - r)
The sum of 2n terms can be calculated using the same formula:
S_2n = a * (1 - r^(2n)) / (1 - r)
We are given that the sum of 2n terms is 33 times the sum of the first n terms:
S_2n = 33 * S_n
Substituting the formulas for S_2n and S_n, we have:
a * (1 - r^(2n)) / (1 - r) = 33 * [a * (1 - r^n) / (1 - r)]
Now, let's simplify the equation:
(1 - r^(2n)) / (1 - r) = 33 * (1 - r^n) / (1 - r)
Cross multiplying, we get:
1 - r^(2n) = 33 * (1 - r^n)
Now, let's isolate the terms with exponents:
1 - r^(2n) = 33 - 33 * r^n
Rearranging the terms, we have:
r^(2n) - 33 * r^n + 32 = 0
This is a quadratic equation in terms of r^n. To solve for r^n, we can substitute u = r^n:
u^2 - 33u + 32 = 0
Factoring the quadratic equation, we get:
(u - 1)(u - 32) = 0
So, u = 1 or u = 32.
Substituting back r^n for u:
r^n = 1 or r^n = 32
If r^n = 1, then r = 1 since any number raised to the power of 0 is 1.
If r^n = 32, then r = 2 since 2^5 = 32.
We need to find the value of n. Let's solve for n in each case:
Case 1: r = 1
If r = 1, then the common ratio is 1 and the geometric progression becomes an arithmetic progression.
Since we are given that the common ratio is 2, this case is not valid.
Case 2: r = 2
If r = 2, then the common ratio is 2.
Substituting this value in the sum formula, we have:
a * (1 - (2^n)) / (1 - 2) = 33 * [a * (1 - 2^n) / (1 - 2)]
Simplifying, we get:
1 - 2^n = 33 - 33 * 2^n
Rearranging the terms, we have:
2^n - 33 * 2^n + 32 = 0
Factoring the quadratic equation, we get:
(1 - 32)(2^n - 1) = 0
So, 2^n - 1 = 0.
Solving for n, we have:
2^n = 1
Taking the logarithm of both sides, we have:
n * log2(2) = log2(1)
n = 0
Therefore, the value of n for which the sum of 2n terms is 33 times the sum of the first n terms is n = 0.
Please let me know if anything is unclear or if you have any further questions!