Chapter 5: Problem 139
Let \(a, b, c\) be the sums of the first \(n\) terms, next \(n\) terms and next n terms of a GP respectively. Then \(a, b, c\) are in (a) \(\mathrm{AP}\) (b) \(\mathrm{GP}\) (c) \(\mathrm{HP}\) (d) AGP
Short Answer
Expert verified
The relationship between a, b, and c, which are sums of consecutive n terms of a geometric progression, is that they form a geometric progression (GP) as well.
Step by step solution
01
Express a, b, and c in terms of A and r
The first term of the GP is A, and the common ratio is r. So the first n terms of the GP are:
A, Ar, Ar^2, Ar^3,..., Ar^(n-1)
The sum of a GP can be found using the formula:
S = A*(1-r^n)/(1-r)
Using this formula, we can find a, b, and c:
a = A*(1-r^n)/(1-r)
b = (Ar^n)*(1-r^n)/(1-r)
c = (Ar^(2*n))*(1-r^n)/(1-r)
02
Check for arithmetic progression (AP)
For a, b, and c to form an AP, there must be a common difference (d) between each consecutive term:
d = b - a
d = c - b
Let's compute d using b and a:
d = (Ar^n)*(1-r^n)/(1-r) - A*(1-r^n)/(1-r)
Factoring A:
d = A[(1-r^n)/(1-r)*r^n - (1-r^n)/(1-r)]
Cancelling out the (1-r^n)/(1-r) term:
d = A*r^n
Now, let's compute the difference between c and b:
d = (Ar^(2*n))*(1-r^n)/(1-r) - (Ar^n)*(1-r^n)/(1-r)
Factoring Ar^n:
d = Ar^n[(1-r^n)/(1-r)*r^n - (1-r^n)/(1-r)]
Cancelling out the (1-r^n)/(1-r) term:
d = Ar^n * r^n
We see that the common difference (d) is different between b and a, and c and b. Hence, a, b, and c do not form an AP.
03
Check for geometric progression (GP)
For a, b, and c to form a GP, there must be a common ratio between each consecutive term:
b/a = c/b
Let's compute b/a and c/b:
b/a = [(Ar^n)*(1-r^n)/(1-r)] / [A*(1-r^n)/(1-r)]
Cancelling out A and (1-r^n)/(1-r) term:
b/a = r^n
c/b = [(Ar^(2*n))*(1-r^n)/(1-r)] / [(Ar^n)*(1-r^n)/(1-r)]
Cancelling out Ar^n and (1-r^n)/(1-r) term:
c/b = r^n
We see that the ratio between b and a, and c and b is the same. Hence, a, b, and c form a GP.
04
Check for harmonic progression (HP)
For a, b, and c to form an HP, the reciprocals of a, b, and c should form an AP. Meaning, 1/a, 1/b, and 1/c should have a common difference.
However, we already found that a, b, and c form a GP. It is not possible for them to also form an HP at the same time. Therefore, a, b, and c do not form an HP.
05
Check for arithmetic geometric progression (AGP)
Since we have already found that a, b, and c form a GP, AGP is not applicable to this problem.
06
Conclusion
We conclude that a, b, and c form a geometric progression (GP). Therefore, the correct answer is (b) \(\mathrm{GP}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Arithmetic Progression
Arithmetic Progression (AP) is a sequence of numbers in which the difference between consecutive terms is constant. This difference is known as the "common difference." In more formal terms, if we have a sequence like:
- a, a+d, a+2d, a+3d, ...,
Harmonic Progression
Harmonic Progression (HP) is a sequence of numbers derived from the reciprocals of an arithmetic progression. If you have an arithmetic progression of the form:
- a, a+d, a+2d, a+3d, ...,
- \(\frac{1}{a}, \frac{1}{a+d}, \frac{1}{a+2d}, \frac{1}{a+3d}, ...\)
Common Ratio
A Geometric Progression (GP) involves a sequence where each term after the first is obtained by multiplying the previous term by a constant called the "common ratio." If we denote this sequence with terms like:
- A, Ar, Ar^2, Ar^3, ...,