Sequences and Series MCQ Quiz - Objective Question with Answer for Sequences and Series - Download Free PDF

Given:

The numbers are 9, 21, and 49, and we need to find the harmonic mean of these three numbers.

Formula used:

The harmonic mean (H) of n numbers a₁, a₂, ..., a_n is given by:

H = n / (1/a₁ + 1/a₂ + ... + 1/a_n)

Calculations:

We have three numbers: 9, 21, and 49. The harmonic mean formula becomes:

H = 3 / (1/9 + 1/21 + 1/49)

First, calculate the reciprocals of the numbers:

1/9 ≈ 0.1111, 1/21 ≈ 0.0476, 1/49 ≈ 0.0204

Now, add the reciprocals:

0.1111 + 0.0476 + 0.0204 = 0.1791

Now, calculate the harmonic mean:

H = 3 / 0.1791 ≈ 16.7

∴ The harmonic mean of 9, 21, and 49 is closest to 16.7.

Given:

The numbers are 27, 60, 108, 150, and 225, and the geometric mean is denoted by x. We need to find the harmonic mean of x and 60.

Formula used:

Geometric mean (x) = (Product of numbers)^(1/n), where n is the number of terms, and the numbers are the given values.

Harmonic mean (H) of two numbers a and b is given by:

H = 2ab / (a + b)

Calculations:

Find the geometric mean (x) of the numbers 27, 60, 108, 150, and 225.

x = (27 × 60 × 108 × 150 × 225)^(1/5)

x = (3 × 3 × 3 × 3 × 2 × 2 × 5 × 3 × 3 × 3 × 2 × 2 × 3 × 5 × 2 × 5 × 3 × 5 × 3 × 5)(1/5)

x = (3¹⁰ × 2⁵ × 5⁵)(1/5)

x = 32 × 2 × 5

x = 90

Find the harmonic mean of x and 60.

Now that we know x = 90, we can find the harmonic mean of x and 60:

H = 2 × 90 × 60 / (90 + 60)

H = 10800 / 150

H = 72

∴ The harmonic mean of x and 60 is 72.

Concept:

​Geometric Progression (G.P.): A sequence where each term is obtained by multiplying the previous term by a constant called the common ratio.

Sum of n terms (S_n) of G.P. is given by: \( S_n = \frac{a(r^n - 1)}{r - 1} \)

• a: First term of the G.P.
• r: Common ratio of the G.P.
• n: Number of terms in the series

Calculation:

Given, f(1) = 3

Using the property: \( f(x + y) = f(x) \times f(y) \)

⇒ f(2) = f(1 + 1) = f(1) × f(1) = 3 × 3 = 9

⇒ f(3) = f(1 + 2) = f(1) × f(2) = 3 × 9 = 27

⇒ f(4) = f(1 + 3) = f(1) × f(3) = 3 × 27 = 81

So, \( f(1), f(2), f(3), \dots \) = 3, 9, 27, 81, … forms a G.P. with a = 3 and r = 3

Given, \( \sum_{x=1}^{n} f(x) = 120 \)

⇒ \( S_n = \frac{3(3^n - 1)}{3 - 1} \)

⇒ \( 120 = \frac{3(3^n - 1)}{2} \)

⇒ \( 120 = \frac{3}{2}(3^n - 1) \)

⇒ \( 240 = 3(3^n - 1) \)

⇒ \( 80 = 3^n - 1 \)

⇒ \( 3^n = 81 \)

⇒ \( 3^n = 3^4 \Rightarrow n = 4 \)

∴ The value of n is 4.

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: Third term of GP = 4

So, a₃ = ar² = 4

Now, product of five time term is given by  = a ×  ar × ar²  × ar³ ×ar⁴   

= a⁵r¹⁰  

= (ar²)⁵ 

= 4⁵

Given:

The sequence P1, P2, P3, ..... is defined by P1 = 211, P2 = 375, P3 = 420, P4 = 523, and Pn = Pn-1 - Pn-2 + Pn-3 - Pn-4 for all n ≥ 5.

Calculation:

We need to find the value of P531 + P753 + P975.

Given the recurrence relation Pn = Pn-1 - Pn-2 + Pn-3 - Pn-4, we notice that this sequence is periodic with a period of 4. This means that every 4 terms, the sequence repeats itself.

Let's calculate the first few terms to identify the pattern:

P1 = 211

P2 = 375

P3 = 420

P4 = 523

P5 = P4 - P3 + P2 - P1 = 523 - 420 + 375 - 211 = 267

P6 = P5 - P4 + P3 - P2 = 267 - 523 + 420 - 375 = -211

P7 = P6 - P5 + P4 - P3 = -211 - 267 + 523 - 420 = -375

P8 = P7 - P6 + P5 - P4 = -375 - (-211) + 267 - 523 = -420

P9 = P8 - P7 + P6 - P5 = -420 - (-375) + (-211) - 267 = -523

We can observe the periodicity:

P1 = 211

P2 = 375

P3 = 420

P4 = 523

P5 = 267

P6 = -211

P7 = -375

P8 = -420

P9 = -523

P10 = 267

The sequence repeats every 4 terms: P5 = 267, P6 = -211, P7 = -375, P8 = -420, P9 = -523, P10 = 267, and so on.

To find P531, P753, and P975, we need to determine their positions in the cycle:

P531 = P(531 mod 4) = P3 = 420

P753 = P(753 mod 4) = P1 = 211

P975 = P(975 mod 4) = P3 = 420

Therefore, P531 + P753 + P975 = 420 + 211 + 420 = 1051

The correct answer is option 1) 898.

Concept:

Arithmetic Progression (AP):

• The sequence of numbers where the difference of any two consecutive terms is same is called an Arithmetic Progression.
• If a be the first term, d be the common difference and n be the number of terms of an AP, then the sequence can be written as follows:
  a, a + d, a + 2d, ..., a + (n - 1)d.
• The sum of n terms of the above series is given by:
  Sn = \(\rm \dfrac{n}{2}[a+\{a+(n-1)d\}]=\left (\dfrac{First\ Term+Last\ Term}{2} \right )\times n\)

Calculation:

The given series is 5 + 9 + 13 + … + 49 which is an arithmetic progression with first term a = 5 and common difference d = 4.

Let's say that the last term 49 is the nth term.

∴ a + (n - 1)d = 49

⇒ 5 + 4(n - 1) = 49

⇒ 4(n - 1) = 44

⇒ n = 12.

And, the sum of this AP is:

S₁₂ = \(\rm \left (\dfrac{First\ Term+Last\ Term}{2} \right )\times 12\)

= \(\rm \left (\dfrac{5+49}{2} \right )\times 12\) = 54 × 6 = 324.

Concept:

Expansion of ex:

\(\rm e^{x} = 1+ \dfrac{x}{1!}+ \dfrac{x^2}{2!}+ \dfrac{x^3}{3!}+ \dfrac{x^4}{4!}+ .....\)

Calculation:

\(\rm e^{x} = 1+ \dfrac{x}{1!}++ \dfrac{x^2}{2!}+ \dfrac{x^3}{3!}+ \dfrac{x^4}{4!}+ .....\)

Put x = -1,

\(\rm e^{(-1)} = 1+ \dfrac{(-1)}{1!}++ \dfrac{(-1)^2}{2!}+ \dfrac{(-1)^3}{3!}+ \dfrac{(-1)^4}{4!}+ .....\)

\( \rm e^{-1} = 1- \dfrac{1}{1!}+ \dfrac{1}{2!}- \dfrac{1}{3!}+ \dfrac{1}{4!}- .....\)

\( \therefore \rm 1- \dfrac{1}{1!}+ \dfrac{1}{2!}- \dfrac{1}{3!}+ \dfrac{1}{4!}- ..... = \frac 1 e\)

Concept:

Five terms in a geometric progression:

If a G.P. has first term a and common ratio r then the five consecutive terms in the GP are of the form \(\rm \dfrac{a}{r^2},\dfrac{a}{r},a,ar,ar^2\) .

Calculation:

Let us consider a general geometric progression with common ratio r.

Assume that the five terms in the GP are \(\rm \dfrac{a}{r^2},\dfrac{a}{r},a,ar,ar^2\).

It is given that third term is 9.

Therefore, a = 9.

Now the product of the five terms is given as follows:

\(\rm \dfrac{a}{r^2}\times\dfrac{a}{r}\times a\times ar \times ar^2 = a^5\)

But we know that a = 9.

Thus, the product is \(9^5=3^{10}\).

Calculation:

\(\rm \frac{1}{{1 \times 2}} + \frac{1}{{2 \times 3}} + \frac{1}{{3 \times 4}} +...+\frac{1}{{n \times (n+1)}} \)

\(\rm = \frac{{2\; - \;1}}{{1 \times 2}} + \frac{{3\; - \;2}}{{2 \times 3}} + \frac{{4\; - \;3}}{{3 \times 4}} +... + \frac{{(n+1)\; - \;n}}{{n \times (n+1)}}\)

\(\rm = \frac{1}{1} - \frac{1}{2} + \;\frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} +... + \frac{1}{n} - \frac{1}{n+1}\)

\(\rm = 1 - \frac {1}{n+1}\)

\(\rm = \frac {n+1 -1}{n+1}\)

\(\rm = \frac {n}{n+1}\)

Concept:

For AP series, 

Sum of n terms  = \(\rm \dfrac n 2\) (First term + nth term)

Calculations:

We know that, For AP series, 

the sum of n terms  = \(\rm \dfrac n 2\) (First term + nth term)

Given, the nth term of the given series is a_n = 5n + 1.

Put n = 1, we get

a1 = 5(1) + 1 = 6.

We know that 

sum of n terms = \(\rm \dfrac n 2\) (First term + nth term)

⇒Sum of n terms = \(\rm \dfrac n 2\) (6 + 5n + 1)

⇒Sum of n terms = \(\rm \dfrac n 2\) (7+ 5n)

Concepts:

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

• Common ratio = r = \(\frac{{{a_2}}}{{{a_1}}} = \frac{{{a_3}}}{{{a_2}}} = \ldots = \frac{{{a_n}}}{{{a_{n - 1}}}}\)
• n^th term of the G.P. is an = arⁿ⁻¹
• Sum of n terms = s = \(\frac{{a\;\left( {{r^n} - 1} \right)}}{{r - 1}}\); where r >1
• Sum of n terms = s = \(\frac{{a\;\left( {1 - {r^n}} \right)}}{{1 - r}}\); where r <1
• Sum of infinite GP = \({{\rm{s}}_\infty } = {\rm{}}\frac{{\rm{a}}}{{1{\rm{}} - {\rm{r}}}}{\rm{}}\); |r| < 1

Where a is 1^st term and r is common ratio.

Calculation:

Given: The third term of a GP is 3

Let 'a' be the first term and 'r' be the common ratio.

∴ T₃ = ar² = 3

We know that T_n = a r^n-1

So, T₁ = a, T₂ = ar, T₃ = ar², T₄ = ar³, T₅ = ar⁴

Concept:

• 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
• 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

Calculation:

Given series is 5, 10, 20, ...

Here, a = 5, r = 2

Sum of n numbers = sn = 1275

To Find: nAs we know that, Sum of n terms of GP = sn = \(\frac{{{\rm{a\;}}\left( {{{\rm{r}}^{\rm{n}}} - 1} \right)}}{{{\rm{r}} - {\rm{\;}}1}}\); where r >1

∴ sn = \(\frac{{{\rm{5\;}}\left( {{{\rm{2}}^{\rm{n}}} - 1} \right)}}{{{\rm{2}} - {\rm{\;}}1}}\)

1275 = 5 × (2n - 1)

⇒ 255 = (2n - 1)

⇒ 2n = 256

⇒ 2n = 28

∴ n = 8

Given :

The harmonic mean and the geometric mean of two numbers are 10 and 12 respectively

Concept used :

(Geometric mean)² = Harmonic mean × Arithmetic mean 

Calculations :

according to the formula given above 

(12)² = 10 × Arithmetic mean 

⇒ Arithmetic mean = 144/10 

⇒ 14.4 

∴ Option 4 will be the correct answer.

Alternate Method 

Concept:

A.M. between a and b = \(a+b \over 2\)

G.M. between a and b = \(\sqrt{ab} \)

H.M. between a and b = \(2ab\over a+b \)

Calculation:

Given, G.M. = 12, H.M. = 10

\(GM = \sqrt{ab} \)

\(12^2 = ab\)

ab = 144 ........(1)

\(HM=\frac{2ab}{a+b}\)

\(10=\frac{2ab}{a+b}\)

2ab = 10a + 10b

ab = 5 (a + b)........(2)

From equation (1) and (2)

144 = 5 (a + b)

\(\frac{144}{5}=a+b\)

\(\frac{144}{10}=\frac{a+b}{2}\)

\(\frac{a+b}{2}=14.4\)

But, we know that,

A.M. between a and b = \(a+b \over 2\)

Therefore, AM = 14.4

Concept:

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

• Common ratio = r = \(\frac{{{a_2}}}{{{a_1}}} = \frac{{{a_3}}}{{{a_2}}} = \ldots = \frac{{{a_n}}}{{{a_{n - 1}}}}\)
• n^th  term of the G.P. is a_n = arⁿ⁻¹
• Sum of n terms = s = \(\frac{{a\;\left( {{r^n} - 1} \right)}}{{r - \;1}}\); where r >1
• Sum of n terms = s = \(\frac{{a\;\left( {{r^n} - 1} \right)}}{{r - \;1}}\); where r <1

Calculation:

Given:

4th term of a G. P is 8 and 10th term is 27

nth  term of the G.P. is T_n = a r^n-1

∴ T₄ = a. r³ = 8      ----(1)

T₁₀ = a r⁹ = 27      ----(2)

Equation (2) ÷ (1), we get 

\( {r^6} = \frac{{27}}{8}\)

\(\Rightarrow {\left( {{{\rm{r}}^2}} \right)^3} = {\rm{\;}}{\left( {\frac{3}{2}} \right)^3}\)

∴ \({r^2} = \frac{3}{2}\)

T₆ = a r⁵

= a r³.r²

\(= 8 \times \frac{3}{2} \)

GIVEN:

Two series given i.e. S1 and S2

FORMULA USED:

a_n = a + ( n - 1 ) d 

S_n = n/2 [2a + (n - 1) d ] 

Where,

a_n = n^th term in the sequence , n= number of terms , a = first term in sequence, d = common difference , Sn = Sum 

CALCULATION:

Here, given series S1  and S2 are in A.P.

So, series will move by adding a fixed common difference ( second term - first term) in consecutive terms 

S1 = 2 , 9 , 16 , 23, 30 , 37 , 44 , 51 ,........ 632      [As  here d = 7 .So, add 7 in previous term to get next term]

S2  = 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51 , ......... 743  [ Here d = 4 ] 

Now, let us take a third series S3 which is common series                    [It will contain common numbers of both series only]

So, from S1 and S2 series, we have 1st common term = 23 , 2nd common term = 51 , d = ( 51 - 23) = 28 

So, add 28 in second term to get third term and so on

S3 = 23 , 51 , .................. \(\le\)  632              [ As, 632 is less than 743 so common between them should be less than 632]

Now, we have a = 23 , d = 28 

⇒ a_n = a + ( n - 1) d    \(\le\) 632 

⇒ 23 + ( n - 1) × 28 \(\le\) 632 

⇒ ( n -  1) × 28  \(\le\)  ( 632 - 23 ) 

⇒ ( n - 1) × 28  \(\le\) 609 

⇒ n -1 \(\le\) 609 /28 

⇒ n - 1 \(\le\) 21.75 

⇒ n \(\le\) 22 .75

 As, n should be equal to or less than 22.75 . So, take n = 22 

Now, as we know that 

Sn = n/2 [ 2a + ( n - 1) d ] 

⇒ Sn = 22/2 [ 2 × 23 + ( 22 - 1) 28]  = 11 [46 + 21 × 28 ] 

⇒ 11 [ 46 + 588 ] = 11 × 634  = 6974

Hence, Sum of all the common terms of the series are 6974 .