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All questions of Harmonic Progressions for RRB NTPC/ASM/CA/TA Exam

If 1/4, 1/x, 1/10 are in HP, then what is the value of x?
  • a)
    5
  • b)
    6
  • c)
    7
  • d)
    8
Correct answer is option 'C'. Can you explain this answer?

Pranab Goyal answered
Given:
1/4, 1/x, 1/10 are in Harmonic Progression (HP)

To find:
The value of x

Solution:

Harmonic Progression (HP):
In a Harmonic Progression (HP), the reciprocals of the terms are in Arithmetic Progression (AP).

Given terms:
1/4, 1/x, 1/10

Reciprocals of the given terms:
4/1, x/1, 10/1

As per the concept of HP:
2 * (1/x) = (1/4) + (1/10)

Solving the equation:
2/x = (10 + 4) / 40
2/x = 14 / 40
x = 40 / 14
x = 20 / 7
x = 7
Therefore, the value of x is 7.
Hence, the correct answer is option C) 7.
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Three numbers 5, p and 10 are in Harmonic progression if p = ?
  • a)
    10/3
  • b)
    20/3
  • c)
    3/10
  • d)
    3/20
Correct answer is option 'B'. Can you explain this answer?

Pranab Goyal answered
Harmonic Progression:
Harmonic progression is a sequence of numbers in which the reciprocals of the terms are in arithmetic progression. In simpler terms, if three numbers a, b, and c are in harmonic progression, then 1/a, 1/b, and 1/c are in arithmetic progression.

Given:
- Three numbers 5, p, and 10 are in harmonic progression.

Calculating the Harmonic Mean (H):
- The Harmonic Mean (H) of three numbers a, b, and c is given by:
H = 3/(1/a + 1/b + 1/c)

Substitute the given numbers:
- Given a = 5, b = p, and c = 10, we can substitute these values into the formula:
H = 3/(1/5 + 1/p + 1/10)

Calculate the Harmonic Mean:
- Simplifying the expression, we get:
H = 3/((10 + 10p + 5)/(10p))
H = 3*(10p)/(15 + 10p)

Given that the numbers are in harmonic progression:
- Since the numbers 5, p, and 10 are in harmonic progression, the Harmonic Mean (H) must be equal to p:
p = 3*(10p)/(15 + 10p)

Solve for p:
- Cross-multiply to solve for p:
p(15 + 10p) = 3*(10p)
15p + 10p^2 = 30p
10p^2 - 15p + 30p = 0
10p^2 + 15p = 0
p(10p + 15) = 0
p = 0 or p = -15/10

Answer:
Therefore, the value of p that satisfies the condition is p = 20/3.

If a, b, c are in geometric progression, then logax x, logbx x and logcx x are in
  • a)
    Arithmetic progression
  • b)
    Geometric progression
  • c)
    Harmonic progression
  • d)
    Arithmetico-geometric progression
Correct answer is option 'C'. Can you explain this answer?

Iq Funda answered
Concept:
If a, b, c are in geometric progression then b2 = ac
If b - a = c - b, than a, b, c are in AP.
If 1/a, 1/b, 1/c are in AP than a, b, c are in HP.
Calculation:
If a, b, c are in geometric progression then b2 = ac
So, by multiplying both side by x2 and taking log on both side to the base x

 are in HP, then which of the following is/are correct?
1. a, b, c are in AP
2. (b + c)2, (c + a)2, (a + b)2 are in GP. Select the correct answer using the code given below.
  • a)
    1 only
  • b)
    2 only
  • c)
    Both 1 and 2
  • d)
    Neither 1 nor 2
Correct answer is option 'A'. Can you explain this answer?

EduRev SSC CGL answered
Concept:
  • If a, b and c are three terms in GP then b2 = ab and vice-versa
  • If three terms a, b, c are in HP, then and vice-versa
  • When three quantities are in AP, the middle one is called as the arithmetic mean of the other two.
  • If a, b and c are three terms in AP then 
  • and vice-versa
  • If a, b and c are in A.P then 1/a, 1/b, 1/c are in H.P and vice-versa
Calculation:
(b + c), (c + a) and (a + b) are in A.P
⇒ 2(c + a) = b + c + a + b    
⇒ 2c + 2a = 2b + a + c     -----(i)
⇒ 2c + 2a - 2b - a - c = 0
⇒ 2b = c + a
So, a, b, c are in AP
Now, Let (b + c)2, (c + a)2, (a + b)2 are in GP
(c + a)2 = √[(b + c).(a + b)]2
⇒ c2 + a2 + 2ac = (b + c).(a + b)
⇒ c2 + a2 + 2ac = ab + b2 + ac + bc
⇒ c2 + a2 - b2 + ac - ab - bc = 0
From here we are unable to check the relation between a, b and c 
So, Our assumption was wrong 
So, (b + c)2, (c + a)2, (a + b)2 are not in GP.

How many two digit numbers are divisible by 7?
  • a)
    13
  • b)
    15
  • c)
    11
  • d)
    None of these
Correct answer is option 'A'. Can you explain this answer?

G.K Academy answered
CONCEPT:
Let us suppose a be the first term and d be the common difference of an AP. Then the nth term of an AP is given by:an = a + (n - 1) × d.
Note: If l is the last term of a sequence, then l = an = a + (n - 1) × d.
CALCULATION:
Here we have to find two digit numbers which are divisible by 7.
i.e 14, 21,............,98 is an AP sequence with first term a = 14, common difference d = 7 and the last term l = 98.
As we know that, if l is the last term of a sequence, then l = an = a + (n - 1) × d
⇒ 98 = 14 + (n - 1) × 7
⇒ 84 = 7(n - 1)
⇒ 12 = n - 1
⇒ n = 13

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