MCQ: Harmonic Progressions - SSC CGL MCQ

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.

Concept:

If a, b, c are in HP then  

If a, b, c are in AP then 2b = a + c

A sequence of numbers is called a Harmonic progression if the reciprocal of the terms are in AP.

Calculation:

Given: 1 / 4, 1 / x, 1 / 10 are in HP

The sequence  1 / 4, 1 / x, 1 / 10 are in HP, so its reciprocals 4, x, 10 are in AP.

As we know, If a, b, c are in AP then 2b = a + c

⇒ 2x = 4 + 10

⇒ 2x = 14

∴ x = 7

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

Concept: 

Three numbers x, y, and z are in H.P if and only if y 

Calculation:

Given: Three numbers 5, p and 10 are in Harmonic progression

Now, According to the concept used

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:a_n = a + (n - 1) × d.

Note: If l is the last term of a sequence, then l = a_n = 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

Given:

n^th term for the first series = n^th term for the second series.

Concept:

Arithmetic Progression: 

• An arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term.
• This fixed number is called the common difference of the AP and it can be positive, negative or zero.

Consider the AP, whose first term is 'a' and common difference is 'd' and there are n number of terms.

a, a + d, a + 2d, a + 3d........a + (n - 1)d

Hence, n^th term of AP is given by

T_n = a + (n - 1)d

Calculation:

Consider the first series

3 + 10 + 17 + ... 

a = 3 and d = 10 - 3 = 7

n^th term of AP is given by,

(T_n)₁ = 3 + (n - 1) × 7 

⇒ (Tn)1 = 7n - 4       .......(1)

Consider the second series 

63 + 65 + 67 + ....

a = 63 and d = 2

nth term of AP is given by,

(Tn)₂ = 63 + (n - 1) × 2

⇒ (Tn)₂ = 2n + 61      .......(2)

According to the question

(Tn)1 = (Tn)2

⇒ 7n - 4 = 2n + 61

⇒ 5n = 65

⇒ n = 13

Hence, for the given series, 13^th term will be equal.

Concept:

Calculation:

From (1), 

∴ a, b and c are in H.P.

Concept:

If a, b, c are in HP then 

If a, b, c are in AP then 2b = a + c

A sequence of numbers is called a Harmonic progression if the reciprocal of the terms are in AP.

Calculation:

Given: 1/2,1/x,1/8 are in HP

The sequence  1/2, 1/x, 1/8 are in HP, so its reciprocals 2, x, 8 are in AP.

As we know, If a, b, c are in AP then 2b = a + c

⇒ 2x = 2 + 8

⇒ 2x = 10

∴ x = 5

Concept:

Let us consider sequence a₁, a₂, a₃ …. an is a G.P.

 

Calculation:

Given: The sequence 2, 6, 18, 54, ..... is a GP.

Here, we have to find which term of the given sequence is 4374.

As we know that, the the general term of a GP is given by: a_n = arⁿ⁻¹

Here, a = 2, r = 3 and let an = 4374

⇒ 4374 = (2) ⋅ (3)n - 1

⇒ 3^n-1 = 2187

⇒ 3n-1 = 3⁷

∴ n - 1= 7

So, n = 8

Hence, 4374 is the 8th term of the given sequence.

Concept:

For any quadratic equation, ax² + bx + c = 0. We have discriminant, D = b² - 4ac, then the given quadratic equation has:

I. Distinct and real roots if D > 0.

II. Real and repeated roots, if D = 0.

III. Complex roots and conjugate of each other, D < 0.

If a, b and c are in HP, then the harmonic mean, 

Calculation:

Given: a (b - c) x² + b (c - a) x + c (a - b) = 0 has equal roots ⇒ Discriminant, D = 0.

By comparing the given equation with the quadratic equation, ax² + bx + c = 0. We get, a’ = a (b - c) , b’ = b (c - a) and c’ = c (a - b)

As, Discriminant, D = 0.

⇒ D = b² - 4ac = b² × (c - a) ² - 4 × a (b - c) × c (a - b) = 0

⇒ D = (bc + ab - 2ac) ² = 0

⇒ bc + ab - 2ac = 0

⇒ b × (a + c) = 2ac

Hence a, b and c are in HP.

Given:

Arithmetic mean = 14

Geometric mean = 12

Formula used:

Calculation:

By using the above formula,

HM = 12²/14

⇒ 144/14 = 72/7

∴ The correct answer is 72/7.

Given that a₁, a₂, a₃,....an are in H.

Let d be the common difference of the A.P., then

adding all the above we get.

Given:

If the first two terms of a H.P. are 2/5 and 12/13 respectively.

Concept:

The harmonic progression for any existing arithmetic progression given as 

a, a+d, a+2d, a+3d, ...... a+(n-1)d is as follows,

Solution:

According to the question,

The a_n for given first term "a" and "d" in general is as follows,

Clearly, 5^th term is the largest.

Concept:

The Harmonic Mean of a & b is

H = 2ab/a+b

Calculation:

let a = 1, b = 3,

H = 2ab/(a+b)

H = (2 × 1 × 3)/(1 + 3)

H = 6/4 = 3/2

Therefore,

Calculation:

The harmonic mean of the the x₁ , x₂ , x₃ ......x_n is given by

The above is geometric mean of common ratio 1/2 and first term 1. The sum of above G.P would be