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If S be the sum, P the product and R the sum of reciprocals of n terms in a GP; then S^n , P , R^-n are in: a. AP b. GP c. HP d. None of these Explain.?
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If S be the sum, P the product and R the sum of reciprocals of n terms...
Answer:
Given: n terms in a GP.
Let the first term be a and the common ratio be r.
Then, the n terms are a, ar, ar^2, ..., ar^(n-1).
1. Sum of n terms in a GP
The formula for the sum of n terms in a GP is given by:
S = (a(r^n - 1))/(r - 1)
2. Product of n terms in a GP
The product of n terms in a GP is given by:
P = a^n * r^(n(n-1)/2)
3. Sum of reciprocals of n terms in a GP
The sum of reciprocals of n terms in a GP is given by:
R = (1/a)*((1 - r^n)/(1 - r))
Now, we need to find the relationship between S^n, P, and R^(-n).
Using the formula for S, we can write:
S^n = [(a(r^n - 1))/(r - 1)]^n
Using the formula for P, we can write:
P = a^n * r^(n(n-1)/2)
Using the formula for R, we can write:
R^(-n) = (1/[(1/a)*((1 - r^n)/(1 - r))])^n
Simplifying the expression for R^(-n), we get:
R^(-n) = [(1 - r^n)/(1 - r)]^n * (1/a)^n
Now, we can see that:
S^n/P = [(a(r^n - 1))/(r - 1)]^n / (a^n * r^(n(n-1)/2))
= [(r^n - 1)/(r - 1)]^n / r^(n(n-1)/2)
= [(r^n - 1)^n / r^n] * [r^(n(n-1)/2) / (r - 1)^n]
= [(1 - 1/r^n)^n / r^n] * [r^(n(n-1)/2) / (1 - r)^n]
= [(1 - 1/r^n)^n / (1 - r^n)] * [r^(n(n-1)/2) / (1 - r)^n]
= (R^(-n))^n
Therefore, we can conclude that S^n/P = R^(-n)^n.
Hence, the given expressions are in a GP. Option (b) is the correct answer.
Given: n terms in a GP.
Let the first term be a and the common ratio be r.
Then, the n terms are a, ar, ar^2, ..., ar^(n-1).
1. Sum of n terms in a GP
The formula for the sum of n terms in a GP is given by:
S = (a(r^n - 1))/(r - 1)
2. Product of n terms in a GP
The product of n terms in a GP is given by:
P = a^n * r^(n(n-1)/2)
3. Sum of reciprocals of n terms in a GP
The sum of reciprocals of n terms in a GP is given by:
R = (1/a)*((1 - r^n)/(1 - r))
Now, we need to find the relationship between S^n, P, and R^(-n).
Using the formula for S, we can write:
S^n = [(a(r^n - 1))/(r - 1)]^n
Using the formula for P, we can write:
P = a^n * r^(n(n-1)/2)
Using the formula for R, we can write:
R^(-n) = (1/[(1/a)*((1 - r^n)/(1 - r))])^n
Simplifying the expression for R^(-n), we get:
R^(-n) = [(1 - r^n)/(1 - r)]^n * (1/a)^n
Now, we can see that:
S^n/P = [(a(r^n - 1))/(r - 1)]^n / (a^n * r^(n(n-1)/2))
= [(r^n - 1)/(r - 1)]^n / r^(n(n-1)/2)
= [(r^n - 1)^n / r^n] * [r^(n(n-1)/2) / (r - 1)^n]
= [(1 - 1/r^n)^n / r^n] * [r^(n(n-1)/2) / (1 - r)^n]
= [(1 - 1/r^n)^n / (1 - r^n)] * [r^(n(n-1)/2) / (1 - r)^n]
= (R^(-n))^n
Therefore, we can conclude that S^n/P = R^(-n)^n.
Hence, the given expressions are in a GP. Option (b) is the correct answer.
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If S be the sum, P the product and R the sum of reciprocals of n terms in a GP; then S^n , P , R^-n are in: a. AP b. GP c. HP d. None of these Explain.?
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If S be the sum, P the product and R the sum of reciprocals of n terms in a GP; then S^n , P , R^-n are in: a. AP b. GP c. HP d. None of these Explain.? for CA Foundation 2024 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about If S be the sum, P the product and R the sum of reciprocals of n terms in a GP; then S^n , P , R^-n are in: a. AP b. GP c. HP d. None of these Explain.? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If S be the sum, P the product and R the sum of reciprocals of n terms in a GP; then S^n , P , R^-n are in: a. AP b. GP c. HP d. None of these Explain.?.
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