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All questions of Geometric Progressions for SSC CGL Exam
If three positive numbers are inserted between 4 and 512 such that the resulting sequence is a G.P., which of the following is not among the numbers inserted?- a)256
- b)16
- c)64
- d)128
Correct answer is option 'D'. Can you explain this answer?
If three positive numbers are inserted between 4 and 512 such that the resulting sequence is a G.P., which of the following is not among the numbers inserted?
a)
256
b)
16
c)
64
d)
128
 | Nilesh Banerjee answered |
Understanding the Problem
To find out which number among the options is not part of the geometric progression (G.P.) inserted between 4 and 512, we need to analyze the properties of a G.P.
Geometric Progression Basics
- A G.P. is a sequence where each term after the first is found by multiplying the previous term by a constant called the common ratio (r).
- For three terms inserted between 4 and 512, the sequence can be expressed as: 4, a, b, c, 512, where a, b, and c are the inserted terms.
Finding the Common Ratio
- The full sequence can be expressed as: 4, 4r, 4r^2, 4r^3, 512.
- Since 512 is the last term, we equate it to 4r^4, giving us the equation: 4r^4 = 512.
Calculating the Common Ratio
- Dividing both sides by 4 gives: r^4 = 128.
- Now, taking the fourth root of both sides yields: r = 2, since 2^4 = 16.
Determining the Inserted Terms
- We can calculate the inserted terms:
- a = 4r = 4 * 2 = 8
- b = 4r^2 = 4 * 2^2 = 16
- c = 4r^3 = 4 * 2^3 = 32
Identifying the Options
The inserted terms are 8, 16, and 32. Now let's evaluate the options:
- a) 256 (not in the sequence)
- b) 16 (in the sequence)
- c) 64 (not in the sequence)
- d) 128 (not in the sequence)
Among the provided options, 128 is not part of the inserted numbers, making option d) 128 the correct answer.
Conclusion
Thus, the number that is not among the inserted terms in the G.P. sequence is 128.
To find out which number among the options is not part of the geometric progression (G.P.) inserted between 4 and 512, we need to analyze the properties of a G.P.
Geometric Progression Basics
- A G.P. is a sequence where each term after the first is found by multiplying the previous term by a constant called the common ratio (r).
- For three terms inserted between 4 and 512, the sequence can be expressed as: 4, a, b, c, 512, where a, b, and c are the inserted terms.
Finding the Common Ratio
- The full sequence can be expressed as: 4, 4r, 4r^2, 4r^3, 512.
- Since 512 is the last term, we equate it to 4r^4, giving us the equation: 4r^4 = 512.
Calculating the Common Ratio
- Dividing both sides by 4 gives: r^4 = 128.
- Now, taking the fourth root of both sides yields: r = 2, since 2^4 = 16.
Determining the Inserted Terms
- We can calculate the inserted terms:
- a = 4r = 4 * 2 = 8
- b = 4r^2 = 4 * 2^2 = 16
- c = 4r^3 = 4 * 2^3 = 32
Identifying the Options
The inserted terms are 8, 16, and 32. Now let's evaluate the options:
- a) 256 (not in the sequence)
- b) 16 (in the sequence)
- c) 64 (not in the sequence)
- d) 128 (not in the sequence)
Among the provided options, 128 is not part of the inserted numbers, making option d) 128 the correct answer.
Conclusion
Thus, the number that is not among the inserted terms in the G.P. sequence is 128.
Find the sum of the series : (20 + 22 + 24 +........+ 28) × 3- a)1023
- b)1331
- c)1024
- d)923
Correct answer is option 'A'. Can you explain this answer?
Find the sum of the series : (20 + 22 + 24 +........+ 28) × 3
a)
1023
b)
1331
c)
1024
d)
923
 | Ssc Cgl answered |
Given:
(20 + 22 + 24 +........+28) × 3
(20 + 22 + 24 +........+28) × 3
Formula Used:
It is a Geometric Progression.
a = first term, r = common ratio
Sum of the Geometric Progression = [a(rn - 1)/(r - 1)]
It is a Geometric Progression.
a = first term, r = common ratio
Sum of the Geometric Progression = [a(rn - 1)/(r - 1)]
Calculation:
a = 1
r = (22/20) = 4/1 = 4
⇒ Sum of the series = [1 × (45 - 1)/(4 - 1)] × 3
⇒ Sum of the series = [1 × (210 - 1)/(3)] × 3
⇒ Sum of the series = [1 × (1024 - 1)]
⇒ Sum of the series = [1 × (1023)]
⇒ Sum of the series = 1023
∴ The Sum of the series is 1023.
a = 1
r = (22/20) = 4/1 = 4
⇒ Sum of the series = [1 × (45 - 1)/(4 - 1)] × 3
⇒ Sum of the series = [1 × (210 - 1)/(3)] × 3
⇒ Sum of the series = [1 × (1024 - 1)]
⇒ Sum of the series = [1 × (1023)]
⇒ Sum of the series = 1023
∴ The Sum of the series is 1023.
The sum of infinite 
- a)25/36
- b)65/20
- c)65/36
- d)100
Correct answer is option 'C'. Can you explain this answer?
The sum of infinite
a)
25/36
b)
65/20
c)
65/36
d)
100
 | EduRev SSC CGL answered |
Given:
Formula used:
Sum of infinite G.P. = a/(1 – r)
Sum of infinite G.P. = a/(1 – r)
Calculation:
Now, multiply both sides by 1/13, we get
⇒ 1/13s = 5/132 + 55/133 ....(2)
Subtracting (2) from (1)
⇒ s – 1/13s = [5/13 + 55/132 + 555/133 + ....] – [5/132 + 55/133 +....]
⇒ 12s/13 = 5/13 + [55/132 – 5/132] + (555/133 – 55/133) + .....
⇒ 12s/13 = 5/13 + 50/132 + 500/133 + ....
Now,
Here R.H.S is an infinite G.P. with first term a = 5/13 and common ratio (r) = 10/13
So,
⇒ 12s/13 = 5/13/(1 – 10/13)
⇒ 12s/13 = 5/13/(13 – 10)/13
⇒ 12s/13 = (5/13)/(3/13)
⇒ 12s/13 = (5/13 × 13/3)
⇒ 12s/13 = 5/3
⇒ s = (13 × 5)/(12 × 3)
⇒ s = 65/36
∴ The required value is 65/36
Now, multiply both sides by 1/13, we get
⇒ 1/13s = 5/132 + 55/133 ....(2)
Subtracting (2) from (1)
⇒ s – 1/13s = [5/13 + 55/132 + 555/133 + ....] – [5/132 + 55/133 +....]
⇒ 12s/13 = 5/13 + [55/132 – 5/132] + (555/133 – 55/133) + .....
⇒ 12s/13 = 5/13 + 50/132 + 500/133 + ....
Now,
Here R.H.S is an infinite G.P. with first term a = 5/13 and common ratio (r) = 10/13
So,
⇒ 12s/13 = 5/13/(1 – 10/13)
⇒ 12s/13 = 5/13/(13 – 10)/13
⇒ 12s/13 = (5/13)/(3/13)
⇒ 12s/13 = (5/13 × 13/3)
⇒ 12s/13 = 5/3
⇒ s = (13 × 5)/(12 × 3)
⇒ s = 65/36
∴ The required value is 65/36
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