A G.P has a common ratio of 2.Find
the value of 'n' for which the sum of
2n terms is 33 times the sum of n
terms?
Well, it seems like this G.P is quite the ambitious one, aiming for 33 times the sum of n terms! Let's figure out the value of 'n' that meets this extravagant goal.
The sum of the n terms in a geometric progression (G.P) is given by the formula:
Sn = a(r^n - 1) / (r - 1)
Where Sn is the sum of the n terms, a is the first term, and r is the common ratio.
Now, the sum of 2n terms would be:
S2n = a(r^(2n) - 1) / (r - 1)
According to the problem, the sum of 2n terms is 33 times the sum of n terms. So, we can set up the equation:
33 * Sn = S2n
Substituting the formulas for Sn and S2n:
33 * [a(r^n - 1) / (r - 1)] = [a(r^(2n) - 1) / (r - 1)]
To make things a bit simpler, let's get rid of the fraction by multiplying both sides by (r - 1):
33 * a(r^n - 1) = a(r^(2n) - 1)
Canceling out the 'a' on both sides:
33 * (r^n - 1) = r^(2n) - 1
Expanding the brackets:
33r^n - 33 = r^(2n) - 1
Moving all the terms to one side to form a quadratic equation:
r^(2n) - 33r^n + 32 = 0
That's quite an equation! Unfortunately, it's not something Clown Bot can solve using humor alone. It requires some mathematical manipulation or the use of a calculator.
To find the value of 'n' for which the sum of 2n terms is 33 times the sum of n terms in a geometric progression (G.P) with a common ratio of 2, we can follow these steps:
Step 1: Write the formula for the sum of n terms in a G.P:
S_n = a * (1 - r^n) / (1 - r), where:
- S_n is the sum of n terms,
- a is the first term,
- r is the common ratio.
Step 2: Find the sum of n terms:
S_n = a * (1 - r^n) / (1 - r)
Step 3: Find the sum of 2n terms:
S_2n = a * (1 - r^(2n)) / (1 - r)
Step 4: The sum of 2n terms is 33 times the sum of n terms:
S_2n = 33 * S_n
Step 5: Substitute the expressions for S_n and S_2n:
a * (1 - r^(2n)) / (1 - r) = 33 * (a * (1 - r^n) / (1 - r))
Step 6: Simplify the equation:
(1 - r^(2n)) / (1 - r) = 33 * (1 - r^n)
Step 7: Multiply both sides by (1 - r):
1 - r^(2n) = 33 * (1 - r) * (1 - r^n)
Step 8: Distribute on the right side:
1 - r^(2n) = 33 + 33 * r^n - 33 * r - 33 * r^(n+1)
Step 9: Rearrange the equation:
r^(n+1) - r^(2n) + 33 * r^n - 33 * r - 32 = 0
Step 10: Solve for 'n' using numerical methods or graphing techniques because it is a nonlinear equation.
To solve this problem, we need to use the formulas for the sum of an n-term geometric progression (G.P).
In a G.P with a common ratio (r), the sum of the first n terms, denoted as Sn, is given by the formula:
Sn = a(1 - r^n) / (1 - r),
where 'a' is the first term and 'n' is the number of terms.
Let's denote the sum of n terms as Sn and the sum of 2n terms as S2n.
Given that the sum of 2n terms is 33 times the sum of n terms, we can write it as an equation:
S2n = 33 * Sn.
Substituting the formulas for Sn and S2n:
a(1 - r^(2n)) / (1 - r) = 33 * [a(1 - r^n) / (1 - r)].
Now, let's simplify this equation step by step:
a(1 - r^(2n)) = 33a(1 - r^n).
Dividing both sides by 'a':
1 - r^(2n) = 33(1 - r^n).
Now, we can solve for 'n'. Let's expand and rearrange the equation:
1 - r^(2n) = 33 - 33r^n.
Move all terms to one side:
- r^(2n) + 33r^n - 32 = 0.
Now, we have a quadratic equation in terms of r^n. To make it easier to solve, let's substitute a variable by letting x = r^n. The equation becomes:
-x^2 + 33x - 32 = 0.
We can solve this quadratic equation for x. Factorizing, we get:
-(x - 32)(x - 1) = 0.
Setting each factor to zero:
x - 32 = 0 or x - 1 = 0.
From these equations, we get two possible values for x:
x = 32 or x = 1.
Now, we substitute back the value of x with r^n:
r^n = 32 or r^n = 1.
If r^n = 32:
This implies that r = (32)^(1/n).
Since the common ratio is known as 2, we can equate the two expressions:
2 = (32)^(1/n).
Taking the nth root of both sides:
2^(n/2) = 32^(1/n).
Since 2^(n/2) = square root of 2^n:
2^(n/2) = (2^5)^(1/n).
Using the properties of exponents, we can simplify further:
2^(n/2) = 2^5/n.
For two powers to be equal, their exponents must be equal:
n/2 = 5/n.
Solving this equation for 'n':
n^2 = 10.
Taking the square root of both sides:
n = sqrt(10) or -sqrt(10).
Since the number of terms cannot be negative, we discard -sqrt(10) as a solution.
Therefore, the value of 'n' for which the sum of 2n terms is 33 times the sum of n terms is n = sqrt(10).