Harmonic Progressions MCQ Quiz - Objective Question with Answer for Harmonic Progressions - Download Free PDF

Explanation:

Let α be the required root.

Sum of roots = \(\frac{-B}{A} = \frac{-b^2(c^2-a^2)}{a^2(b^2-c^2)}\)

⇒ α + α = \( = \frac{-b^2(c^2-a^2)}{a^2(b^2-c^2)}\)

⇒ α = \( \frac{b^2(c^2-a^2)}{2a^2(c^2-b^2)}\)

∴ Option (c) is correct

Explanation:

Given:

 ⇒ a² (b² – c ² ) x² + b ² (c ² – a ² ) x + c ² (a ² – b ² ) = 0...(i)

Now  Equation has equal roots:

Then D =0

⇒ [b² (c ² – a² )]² – 4a ² c ² (b ² – c ² ) (a ² – b ² ) = 0

⇒  b⁴ (c ² – a ² )² – 4a ² c ² (b ² – c ² ) (a ² – b ² ) = 0

Solving above equation, we get 

⇒  b⁴ (c² + a² )² = 4a ⁴ c ⁴ 

⇒ b ² (a ² + c ² ) = 2a ² c ²

⇒ \(\frac{2}{b^2} = \frac{1}{a^2}+ \frac{1}{c^2}\)

⇒ a² , b ² and c ² are in HP

∴ Option (c) is correct.

Explanation:

Given,

x, y, z are in GP

Thus, 

⇒ y² =xz

⇒ \(2lny = lnx + lnz\)

lnx, ln y and lnz are in AP

⇒ (1 + lnx), (1 + lny) and (1 + lnz) are in AP

⇒ \(\frac{1}{lnx}, \frac{1}{lny}, \frac{1}{lnz}\) are in HP

Calculation

given a₁, a₂, a₃ .... are in H.P.

Therefore, \(\frac{1}{a_{1}}, \frac{1}{a_{2}}, \frac{1}{a_{3}}\), .... are in A.P.

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

∴ \(\frac{1}{a_{n}}=\frac{1}{a_{1}}+(n-1) d \text { and } \frac{1}{a_{20}}=\frac{1}{a_{1}}+19 d\)

⇒ \(\frac{1}{a_{n}}=\frac{1}{5}+(n-1) d \text { and } \frac{1}{25}=\frac{1}{5}+19 d\)

⇒ \(\frac{1}{a_{n}}=\frac{1}{5}+(n-1) d \text { and } d=\frac{-4}{19 \times 25}\)

⇒ \(\frac{1}{a_{n}}=\frac{1}{5}-\frac{4(n-1)}{19 \times 25}\)

⇒ \(\frac{1}{a_{n}}=\frac{95-4 n+4}{19 \times 25}\)

⇒ \(a_{n}=\frac{99-4 n}{19 \times 25}\)

Now, a_n < 0

⇒ \(\frac{99-4 n}{19 \times 25}<0\)

⇒ 90 - 4n < 0

⇒ 4n > 99

⇒ \(n>24 \frac{3}{4} \Rightarrow n \geq 25\)

Hence option 4 is correct

Concept:

Using AM-GM-HM inequality, AM ≥ GM ≥ HM

Equality holds when all when all numbers are equal.

Calculation:

Given, a, b, c, d are in HP

⇒ b is the HM between a and c, and c is HM between b and d.

AM between a and c = \(\frac{1}{2}\)(a + c)

AM between b and d = \(\frac{1}{2}\)(b + d)

Now, as AM > HM

⇒ a + c > 2b …(i) 

Also, c is HM between b and d

⇒ b + d > 2c …(ii)

Adding (i) and (ii), we get:

(a + c) + (b + d) > 2(b + c)

⇒ a + d > b + c 

∴ If a, b, c, d ∈ R+ and a, b, c, d are in H.P., then b + c < a + d.

The correct answer is Option 4.

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, \(b = \frac{{2ac}}{{a + c}}\)

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

\(\Rightarrow b=\frac{2ac}{a+c}\)

Concept:

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

• Common ratio = r = \(\frac{{{{\rm{a}}_2}}}{{{{\rm{a}}_1}}} = \frac{{{{\rm{a}}_3}}}{{{{\rm{a}}_2}}} = \ldots = \frac{{{{\rm{a}}_{\rm{n}}}}}{{{{\rm{a}}_{{\rm{n}} - 1}}}}\)
• nth  term of the G.P. is an = arn−1

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: \(\rm {a_n} = a{r^{n\; - \;1}}\)

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.

Additional Information

• Sum of n terms of GP = sn = \(\frac{{{\rm{a\;}}\left( {{{\rm{r}}^{\rm{n}}} - 1} \right)}}{{{\rm{r}} - {\rm{\;}}1}}\); where r >1
• Sum of n terms of GP = sn = \(\frac{{{\rm{a\;}}\left( {1 - {\rm{\;}}{{\rm{r}}^{\rm{n}}}} \right)}}{{1 - {\rm{\;r}}}}\); where r <1
• Sum of infinite GP = \({{\rm{s}}_\infty } = {\rm{\;}}\frac{{\rm{a}}}{{1{\rm{\;}} - {\rm{\;r}}}}{\rm{\;}}\) ; |r| < 1

Concept:

If a, b, c are in HP then \(\rm b = \frac{2ac}{a+c}\)

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: \(\rm \frac 1 2, \frac 1 x, \frac 1 {8}\) are in HP

The sequence  \(\rm \frac 1 2, \frac 1 x, \frac 1 {8}\) 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:

If a, b, c are in A.P ⇔ \( \rm b =\frac{a+c}{2}\)

If a, b, c are in G.P ⇔  b² = ac

If a, b, c are in H.P ⇔ \(\rm b =\frac{2ac}{a+c} \)   

Calculation:

Given: \(\rm x^a=y^b=z^b\) and x, y and z are in GP

∴ y² = xz         ....(1)

Let, \(\rm x^a=y^b=z^c =k\)

\(\rm ⇒ x=(k)^{\frac1a},\ y=(k)^{\frac1b},\ z=(k)^{\frac1c}\)

From (1), 

\(\rm (k)^{\frac2b}= (k)^{\frac1a}\times(k)^{\frac1c}\)

\(\rm (k)^{\frac2b}= (k)^{\frac{1}{a}+\frac{1}{c}}\)

\(⇒ \rm \frac{2}{b}=\frac1a+ \frac1c\)

\(⇒ \rm \frac{2}{b}= \frac{a+c}{ac}\)

\(\Rightarrow \rm b =\frac{2ac}{a+c} \)

∴ a, b and c are in H.P.

Hence, option (3) is correct. 

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.

Additional Information

Sum of n terms of series, whose first term is a and the common difference is d.

\(S_n = \frac{n}{2}[ 2a + (n - 1)d]\)

CONCEPT : 

Three numbers x, y, and z are in H.P if and only if y = \(\frac{2xz}{x+z}\)

So, \(\rm {1\over x}\ , {1\over y} \ and \ {1\over z}\) are in A.P. if and only if: \(\frac{2}{y}=\frac{1}{x}+\frac{1}{z}\)

CALCULATION:

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

⇒ \(\rm {1\over 5}\ , {1\over p} \ and \ {1\over 10}\) are in A.P

Now, According to the concept used

⇒ 2 ⋅\(\rm {1\over p}\) = \(\rm {1\over 5} + \rm {1\over 10}\)

⇒ 2 ⋅\(\rm {1\over p}\) = \(\rm \frac {2\ + \ 1}{10}\)

⇒ \(\frac{2}{p}=\frac{3}{10}\)

⇒ p = \(\rm 20\over 3\)

∴ The value of p is 20/3.

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

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.

\({\log _a}b = \frac{1}{{{{\log }_b}a}}\)

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

\({\log _x}({x^2}{b^2}) = {\log _x}({x^2}ac)\)

\({\log _x}({x^2}{b^2}) = {\log _x}(xa\cdot xc)\)

\(2{\log _x}xb = {\log _x}xa + {\log _x}xc\)

\({\log _x}xb - {\log _x}xa = {\log _x}xc - {\log _x}xb\)

so logx ax, logx bx and logx cx are in AP.

\(\frac{1}{{{{\log }_{ax}}x}},\frac{1}{{{{\log }_{bx}}x}},\frac{1}{{{{\log }_{cx}}x}}\) are also in AP

\({\log _{ax}}x,{\log _{bx}}x,{\log _{cx}}x\) are in HP

Concept:

The harmonic mean between two numbers a and b is given by: 

HM = \(\rm \frac{2ab}{a+b}\)

Calculation:

Given:  \(\rm \frac {a^{n + 1} + b^{n + 1}}{a^n + b^n}\) be the harmonic mean of a and b

To Find: n

As we know, 

The harmonic mean between two numbers a and b is given by: 

HM = \(\rm \frac{2ab}{a+b}= \rm \frac {a^{n + 1} + b^{n + 1}}{a^n + b^n}\)

\(\Rightarrow \rm 2ab \times (a^n + b^n) = (a+b)(a^{n + 1} + b^{n + 1}) \\ \Rightarrow 2a^{n + 1}b + 2a^nb^{n + 1} = a^{n + 2} + b^{n + 2}+a^{n + 1}b + 2a^nb^{n + 1} \\ \Rightarrow a^{n + 2} + b^{n + 2} - a^{n + 1}b - a^nb^{n + 1} =0 \\ \Rightarrow a^{n + 1}(a-b) -b^{n + 1}(a-b)=0 \\ \Rightarrow (a^{n + 1}-b^{n + 1}) (a-b) = 0 \\\therefore (a^{n + 1}-b^{n + 1}) = 0\)

This is possible only when n + 1 = 0

Therefore, n = -1

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 b = \(\rm \frac{2ac}{a + c}\)  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 b = \(\rm \frac{a + c}{2}\)  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:

\(\frac{1}{b+c}, \frac{1}{c+a},\frac{1}{a+b}\) are in HP

(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.

∴ Only the 1st option is correct.