CRE 3 - Infinite Geometric Progression | Algebra - Progressions
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The sum of an infinite G.P. can be found
- (a)
For all values of r
- (b)
For only positive value of r
- (c)
Only for 0 < r < 1
- (d)
Only for – 1 < r < 1
Answer: Option D
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Explanation :
S∞ = a/(1 - r) where -1 < r < 1 i.e. |r| < 1
Hence, option (d).
Workspace:
Find the sum of the following infinite series: 3 + 1 + 1/3 + 1/9 + …
- (a)
5
- (b)
6
- (c)
9
- (d)
4.5
Answer: Option D
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Explanation :
We know the sum of an infinite GP = a/(1 - r)
where, a is the first term and r is the common ratio and -1 < r < 1.
Here, a = 3 and r = 1/3.
∴ S∞ = 3/(1 - 1/3) = 9/2 = 4.5.
Hence, option (d).
Workspace:
If 3 + 3α + 3α2 + … ∞ = 15/4, then the value of α will be
- (a)
15/23
- (b)
1/5
- (c)
7/8
- (d)
15/7
Answer: Option B
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Explanation :
Given, 3 + 3α + 3α2 + 3α3 + … ∞ = 15/4
We know the sum of an infinite GP = a/(1-r)
⇒ 3[1/(1 - α)] = 15/4
⇒ 4 = 5(1 - α)
⇒ α = 1/5
Hence, option (b).
Workspace:
The sum of infinity of a geometric progression is 16 and the first term is 9. The common ratio is
- (a)
7/16
- (b)
9/16
- (c)
1/9
- (d)
7/9
Answer: Option A
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Explanation :
We know the sum of an infinite GP = a/(1 - r)
∴ 16 = 9/(1 - r)
⇒ r = 7/16
Hence, option (a).
Workspace:
The mid-points of the sides of a triangle are joined forming a second triangle. Again, a third triangle is formed by joining the mid-points of this second triangle and this process is repeated infinitely. If the perimeter and area of the outer triangle are P and A respectively. What will be the sum of perimeters of all the triangles?
- (a)
2P
- (b)
P2
- (c)
3P
- (d)
P2/2
Answer: Option A
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Explanation :
The length of sides of successive triangles will form a G.P. with common ratio 1/2. [Basic Proportionality Theorem]
∴ Total perimeter = P + P/2 + P/4 + … ∞
⇒ Total perimeter = P/(1 - 1/2) = 2P
Hence, option (a).
Workspace:
In the above problem, find the sum of areas of all the triangles
- (a)
4A/5
- (b)
4A/3
- (c)
3A/4
- (d)
5A/4
Answer: Option B
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Explanation :
Since, the side of each successive triangle is half, area of each successive triangle will be 1/4th i.e., the area of successive triangles will form an infinite G.P. with a common ratio 1/4.
∴ Total Area = A + A/4 + A/ 16 + … ∞
⇒ Total area = A/(1 - 1/4) = 4A/3
Hence, option (b).
Workspace:
In an infinite G.P. each term of the GP is equal to the sum of all the following terms. Find the ratio of this G.P.
- (a)
1/2
- (b)
1/3
- (c)
1/4
- (d)
Cannot be determined
Answer: Option A
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Explanation :
Let the first term be a and the common ratio be r.
Given, each term of the the GP is twice the sum of all the following terms.
Let’s consider the 3rd term of the G.P.
∴ ar2 = ar3 + ar4 + ar5 + … ∞
⇒ ar2 = (ar3)/(1-r)
⇒ 1 – r = r
⇒ r = ½
Hence, option (a).
Workspace:
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