Arithmetic Mean MCQ Quiz in বাংলা - Objective Question with Answer for Arithmetic Mean - বিনামূল্যে ডাউনলোড করুন [PDF]

CONCEPT : 

If three numbers x, y, and z are in arithmetic progression then 2y = x + z.

CALCULATION:

Given: Three numbers 5, p and 11 are in arithmetic progression

As we know that, If three numbers x, y, and z are in arithmetic progression then 2y = x + z.

⇒ 2p = 5 + 11

⇒ 2p = 16

⇒ p = 8

Hence, option (1) is the correct answer.

Concept:

For any positive real numbers a1, a2, a3, ….., an. The following relation holds:

A.M. ≥ G.M.    

•  A.M. = \(\frac{a_{1}+a_{2}+a_{3}+.....+a_{n}}{n}\)      ----(1)

• G.M. = \(\sqrt[n]{a_{1}a_{2}a_{3}.....a_{n}}\)      ----(2)

Calculation:

For the maximum value of a³b, there should be four terms among which three terms are containing variable a and one term containing the variable b. So that, when we multiply the terms, we get a3b. We have to do all this without distorting the equation given in the problems.

⇒ The terms of the equation would be a, a, a, and 3b.

Since A.M. ≥ G.M.    

\(⇒ \frac{a+a+a+3b}{4}≥ \sqrt[4]{a\times a\times a\times 3b}\)      [using (1) & (2)]

⇒ (3a + 3b)/4 ≥ (3a³b)^1/4

⇒ [(3a + 3b)/4]⁴ ≥ 3a3b

⇒ (3/4)⁴ (a + b)⁴ ≥ 3a3b

⇒ (81/256) (a + b)4 ≥ 3a3b

⇒ 27/256 ≥ \(\dfrac{a^3b}{(a + b)^4}\)

Hence, the maximum value of \(\dfrac{a^3b}{(a + b)^4}\) is \(\dfrac{27}{256}\)

Given:

Two numbers: x = a - b and y = a + b

Concept used:

The arithmetic mean (average) between two numbers x and y is given by:

Arithmetic Mean = \(\dfrac{x + y}{2}\)

Calculation:

Using the given values of x and y:

Arithmetic Mean = \(\dfrac{(a - b) + (a + b)}{2}\)

Simplifying the expression in the numerator:

Arithmetic Mean = \(\dfrac{2a}{2}\)

Canceling out the common factor of 2:

Arithmetic Mean = a

∴ The arithmetic mean between a - b and a + b is a.

Calculation

Given

log_ea, log_eb, log_ec  are in A.P. 

⇒ b² = ac   …..(i) 

Also \(\log _c\left(\frac{\mathrm{a}}{2 \mathrm{~b}}\right),\log _{\mathrm{e}}\left(\frac{2\mathrm{~b}}{3\mathrm{c}}\right),\log _{\mathrm{e}}\left(\frac{3\mathrm{c}}{\mathrm{a}}\right) \) are in A.P. 

⇒ \(\left(\frac{2\mathrm{~b}}{3\mathrm{c}}\right)^2=\frac{\mathrm{a}}{2\mathrm{~b}}\times\frac{3\mathrm{c}}{\mathrm{a}} \)

⇒ \( \frac{\mathrm{b}}{\mathrm{c}}=\frac{3}{2}\)

Putting in eq. (I)

⇒ \(\mathrm{b}^2=\mathrm{a} \times \frac{2 \mathrm{~b}}{3}\)

⇒ \(\frac{\mathrm{a}}{\mathrm{b}}=\frac{3}{2}\)

a ∶ b ∶ c = 9 ∶ 6 ∶ 4 

Hence option (1) is correct

Given:

The arithmetic mean of five numbers is zero. The numbers may not be distinct. 

Concept:

Apply concept of mean and sum of numbers.

Calculation:

The arithmetic mean of five numbers being zero implies that the sum of the numbers is zero.

Let the five numbers be a, b, c, d, e.

\(\rm \frac{a+b+c+d+e}{5}=0\)

\(\rm a+b+c+d+e=0\)

Now, let's consider each option:

(1) The product of the numbers is zero:

This is not necessarily true. The sum being zero does not imply that the product is zero. For example, 

1+1+1+1+(−4)=0, but the product is not 0

(2) At most two of these numbers are positive:

This is not necessarily true. Consider the example 

1+1+1+1+(−4)=0. In this case,four numbers are positive.

(3) There cannot be exactly one zero:

This is not necessarily true. Consider the example 

0+1+1+(-1)+(-1)=0. In this case, one is exactly zero.

(4) There cannot be exactly one non-zero number:

Hence the option (4) is correct. 

Concept:

Mean: Let X be a random variable with values x_i for i = 1, 2, …, n with frequencies f_i. So:

Mean \(= \frac{{\mathop \sum \nolimits_{{\rm{i}} = 0}^{\rm{n}} {{\rm{x}}_{\rm{i}}}{{\rm{f}}_{\rm{i}}}}}{{\mathop \sum \nolimits_{{\rm{i}} = 0}^{\rm{n}} {{\rm{f}}_{\rm{i}}}}}\)

Binomial expansion formula:

• (1+ x)ⁿ = ⁿC₀x⁰ + ⁿC₁x¹ + ⁿC₂x² + … + ⁿC_nxⁿ

Calculation:

Given: 

x_i = i for i = 0, 1, 2, …, n

f_i = ⁿC_i for i = 0, 1, 2, …, n

So, mean \(= \frac{{\mathop \sum \nolimits_{{\rm{i}} = 0}^{\rm{n}} {{\rm{x}}_{\rm{i}}}{{\rm{f}}_{\rm{i}}}}}{{\mathop \sum \nolimits_{{\rm{i}} = 0}^{\rm{n}} {{\rm{f}}_{\rm{i}}}}}\)

\(= \frac{{0 \times {{\rm{\;}}^{\rm{n}}}{{\rm{C}}_0} + 1 \times {{\rm{\;}}^{\rm{n}}}{{\rm{C}}_1} + 2 \times {{\rm{\;}}^{\rm{n}}}{{\rm{C}}_2} + \ldots + {\rm{n}} \times {{\rm{\;}}^{\rm{n}}}{{\rm{C}}_{\rm{n}}}}}{{{{\rm{\;}}^{\rm{n}}}{{\rm{C}}_0} + {{\rm{\;}}^{\rm{n}}}{{\rm{C}}_1} + {{\rm{\;}}^{\rm{n}}}{{\rm{C}}_2} + \ldots + {{\rm{\;}}^{\rm{n}}}{{\rm{C}}_{\rm{n}}}}}\)

Now we know that,

(1+ x)ⁿ = ⁿC₀x⁰ + ⁿC₁x¹ + ⁿC₂x² + … + ⁿC_nxⁿ ----(1)

Differentiating both sides w.r.t x

⇒ n(1 + x)^n-1 = 0 + 1 × ⁿC₁x + 2 × ⁿC₂x¹ + … + n × ⁿC_nxⁿ ----(2)

Put x = 1 in equation (2)

⇒ n × 2^n-1 = 0 + 1 × ⁿC₁ + 2 × ⁿC₂ + … + n × ⁿC_n ----(3)

Put x = 1 in equation (1)

⇒ 2ⁿ = ⁿC₀x⁰ + ⁿC₁x¹ + ⁿC₂x² + … + ⁿC_nxⁿ ----(4)

So, \(\frac{{0 \times {{\rm{\;}}^{\rm{n}}}{{\rm{C}}_0} + 1 \times {{\rm{\;}}^{\rm{n}}}{{\rm{C}}_1} + 2 \times {{\rm{\;}}^{\rm{n}}}{{\rm{C}}_2} + \ldots + {\rm{n}} \times {{\rm{\;}}^{\rm{n}}}{{\rm{C}}_{\rm{n}}}}}{{{{\rm{\;}}^{\rm{n}}}{{\rm{C}}_0} + {{\rm{\;}}^{\rm{n}}}{{\rm{C}}_1} + {{\rm{\;}}^{\rm{n}}}{{\rm{C}}_2} + \ldots + {{\rm{\;}}^{\rm{n}}}{{\rm{C}}_{\rm{n}}}}} = \frac{{{\rm{n}} \times {2^{{\rm{n}} - 1}}}}{{{2^{\rm{n}}}}}\) [Using equation (3) & (4)]

\(= \frac{{\rm{n}}}{2}\)

Concept:

If third order equation ax³ + bx² + cx + d =0 has a root α, β and γ then;

\(α + β + γ = - \frac{b}{a}\)

\(α β + β γ + γ α = \frac{c}{a}\)

\(α β γ = - \frac{d}{a}\)

Calculation:

Given:

x3 - 12x2 + 39x - 28 = 0

Let roots of given equation is α – d, α, α + d

\(α - d + α + α + d = - \frac{{ - 12}}{1}\)

3 α = 12

α = 4

\(\left( {α - d} \right)\left( α \right)\left( {α + d} \right) = - \left( {\frac{{ - 28}}{1}} \right)\)

4 – d ⋅ 4 ⋅ 4 + d = 28

16 – d² = 7

D² = 16 - 7 = 9

D = 3

Concept:

• When any two consecutive numbers in a series have the same difference, they are in Arithmetic Progression or AP. An example of AP is 1 , 3 , 5 ...
• The difference between these two consecutive numbers of the series is called the common difference and it is often denoted by d.
• The formula for the arithmetic mean of two numbers a and b is \(A=\dfrac{a+b}{2}\)
• If there are n arithmetic means between two numbers a and b, then the numbers and the arithmetic means of the numbers are in AP where the first number is a and the last number is b.
• The sum of n numbers in AP is \(S_n=\dfrac{n}{2}(a+b)\) where a is the first and b is the last number of the AP.

Calculation:

Given, A is the arithmetic mean between two numbers and S is the sum of n arithmetic means between the same numbers.

Let a and b be the two numbers in question. Then the according to the formula, the arithmetic mean between the two numbers \(A=\dfrac{a+b}{2}\)

Again, according to the formula, the sum of n arithmetic means between the same numbers a and b is \(S=\dfrac{n}{2}(a+b)\) as all the n arithmetic means are in AP.

Hence, the relation between A and S is \(S=nA\)

Concept:

If A is the arithmetic mean between a and b, 

\(⇒ A=\frac{a+b}{2}\)

If G is the geometric mean between a and b,

⇒ G = √ab

The quadratic equation having roots (α, β) is x² - (α + β)x + αβ = 0.

Calculation:

We have, the arithmetic mean of a and b is 10.

\(⇒ 10=\frac{a+b}{2}\)

⇒ a + b = 20      ----(1)

Also, the geometric mean of a and b is 8.

⇒ 8 = √ab

⇒ ab = 64      ----(2)

Clearly, from equations (1) & (2), we get that a and b are the roots of the quadratic equation given below:

x² - (a + b)x + ab = 0

⇒ x2 - 20x + 64 = 0

⇒ (x - 16)(x - 4) = 0

⇒ x = 16 or 4

⇒ a = 16 and b = 4 or a = 4 or b = 16

Hence, the numbers are 4 and 16.

Concept:

If three terms a, b, c are in G.P.,

then, b is the geometric mean of a and c

that is, b² = ac or b = √ac .

AM of a and b = \({a + b\over 2}\)

Calculation:

Given, a, b, c are in GP

⇒ b = √ac 

And, x, y are the AM's between a, b and b, c respectively

⇒ x = \({a + b\over 2}\) and y = \({b + c\over 2}\)

⇒ x + y = \({a + b\over 2}\) + \({b + c\over 2}\)

⇒ x + y = \({a\ +\ 2b\ + \ c\over 2}\)

⇒ x + y = \({a\ +\ 2\sqrt {ac}\ + \ c\over 2}\)

⇒ x + y = \({(\sqrt a)^2\ +\ 2\sqrt a \sqrt c\ + \ (\sqrt c)^2\over 2}\)

⇒ x + y = \(\frac{(\sqrt{a}+\sqrt{c})^2}{2}\)

So, only statement I is correct.

∴ The correct answer is option (1).