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

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.