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.
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.
1 answer