Progression MCQ Quiz - Objective Question with Answer for Progression - Download Free PDF

Given:

The third term is greater than the first by 9.

The second term is greater than the fourth by 18.

Concept:

n^th Term of GP whose first term & common ratio are 'a' & 'r', is given by T_n = ar^n-1

Calculation:

Let a be the first term and r be the common ratio of the G.P.

a₁ = a, a₂ = ar, a₃ = ar², a₄ = ar³

According to the question

a₃ = a₁ + 9

⇒ ar² = a + 9 ....(1)

Now, a₂ = a₄ + 8

⇒ ar = ar³ + 18 ....(2)

From equation (1) and (2),

⇒ a(r² - 1) = 9 ....(3)

⇒ ar(1 - r²) = 18 ...(4)

Dividing (4) and (3), we get

⇒ \(\frac{ar(1-r^2)}{{a(r^2-1)}}\) = \(\frac{18}{{9}}\)

⇒ -r = 2

⇒ r = -2

Substituting the value of r in equation (1), we get

⇒ 4a = a + 9

⇒ 3a = 9

⇒ a = 3

So. the first term of G.P. a₁ = a = 3

∴ The required value is 3.

Given:

N₁ = 3 + 33 + 333 + .... + 333333 and

N₂ = 4 + 44 + 444 + .... + 4444444

Calculations:

N₁ = 3 + 33 + 333 + 3333 + 33333 + 333333

⇒ N₁ = 3 × (1 + 11 + 111 + 1111 + 11111 + 111111)

Sum of series: 1 + 11 + 111 + 1111 + 11111 + 111111 = 123456

⇒ N₁ = 3 × 123456 = 370368

N₂ = 4 + 44 + 444 + 4444 + 44444 + 444444 + 4444444

⇒ N₂ = 4 × (1 + 11 + 111 + 1111 + 11111 + 111111 + 1111111)

Sum of series: 1 + 11 + 111 + 1111 + 11111 + 111111 + 1111111 = 1234567

⇒ N₂ = 4 × 1234567 = 4938268

N₁ + N₂ = 370368 + 4938268

⇒ N₁ + N₂ = 5308636

Sum of digits of 5308636 ⇒ 5 + 3 + 0 + 8 + 6 + 3 + 6 = 31

The correct answer is option 2.

Concept Used:

The Geometric Mean (G.M) of a series containing n observations is the nth root of the product of the values.

\(\begin{array}{l}G. M = \sqrt[n]{x_{1}× x_{2}× …x_{n}}\end{array}\)

Calculation:

Using the above formula -

⇒ 8⁵ = 32 × 4 × 8 × X × 2

⇒ X =  \(\frac{8^{5}}{32\times 4\times 8\times 2}\)= 16

∴ The correct answer is 16

Given:

Two APs have the same common difference. The first term of one of these is –1 and that of the other is – 8.

Concept:

For any given AP such that its first term is ' a ' and the common difference is ' d '

a_n = a + (n - 1)d

Solution:

According to the question, the common difference of the two APs is the same,

Let's say the common difference is ' d '.

For the first AP

The first term is -1 and the common difference is ' d '

The fourth term will be,

m₄ = -1 + (4 - 1)d = -1 + 3d

For the second AP

The first term is -8 and the common difference is ' d '

The fourth term will be,

n4 = -8 + (4 - 1)d = -8 + 3d

The difference between the 4^th term is as follows,

m₄ - n₄ = -1 + 3d - ( -8 + 3d )

m4 - n4 = 7

Hence, option 3 is correct.

Given:

The fifth term of a geometric progression (GP) = 27

The eighth term of the GP = 729

Formula used:

General term of GP: T_n = a × r^n-1

Where, a = first term, r = common ratio, n = term number

Calculation:

Fifth term: T₅ = a × r⁴ = 27

Eighth term: T₈ = a × r⁷ = 729

Dividing both equations:

⇒ (a × r⁷) / (a × r⁴) = 729 / 27

⇒ r³ = 27

⇒ r = 3

Substitute r = 3 into T₅:

⇒ a × 3⁴ = 27

⇒ a × 81 = 27 

⇒ a = 27 / 81 = 1/3

T₁₁ = (1/3) × 3¹⁰

⇒ T₁₁ = (1/3) × 59049

⇒ T₁₁ = 19683

∴ The correct answer is option (1).

Formula used:

Sum of geometric progression (S_n) = {a × (rⁿ - 1)}/(r - 1)

Where, a = first term ; r = common ratio ; n = number of term

Calculation:

3 +32 + 33 +...+ 38.

Here, a = 3 ; r = 3 ; n = 8

Sum of the series (S₈) = {a × (r8 - 1)}/(r - 1)

⇒ {3 × (3⁸ - 1)}/(3 - 1)

⇒ (3 × 6560)/2 = 3280 × 3 

⇒ 9840

∴ The correct answer is 9840.

Given:

13 + 23 + …….. + 93 = ?

Formula:

S_n = n/2 [a + l]

T_n = a + (n – 1)d

n = number of term

a = first term

d = common difference

l = last term

Calculation:

a = 13

d = 23 – 13 = 10

T_n = [a + (n – 1)d]

⇒ 93 = 13 + (n – 1) × 10

⇒ (n – 1) × 10 = 93 – 13

⇒ (n – 1) = 80/10

⇒ n = 8 + 1

⇒ n = 9

S₉ = 9/2 × [13 + 93]

= 9/2 × 106

= 9 × 53

= 477

Formula used:

an = a + (n – 1)d

Here, a → first term, n → Total number, d → common difference, an → n^th term

Calculation:

First three-digit number divisible by 6, (a) = 102 

Last three-digit number divisible by 6, (an) = 996

Common difference, (d) = 6 (Since the numbers are divisible by 6)

Now, a_n = a + (n – 1)d

⇒ 996 = 102 + (n – 1) × 6 

⇒ 996 – 102 = (n – 1) × 6

⇒ 894 = (n – 1) × 6

⇒ 149 = (n – 1)

⇒ n = 150

∴ The total three digit number divisible by 6 is 150

Formula used:

S_n = [n x (a + a_n) ] /2

a_n = a + (n-1)d

d = difference

a = initial term

a_n = last term

n = number of terms

S_n = Sum of n terms

Solution:

The series can be written as:

 \(1\over99\)[99x99+11 + 99x99+13 + ... + 99x99+67]

= \(1\over99\) [9812 + 9814 + 9816+ ... + 9868]

 Now, our series is, 9812, 9814,...,9868.

a = 9812

a_n = 9868

d = 9814 - 9812 = 2

9868= 9812 + (n-1) x 2

n - 1 = 56/2 = 28

n = 29

S_n = 29 x (9812 + 9868) / 2 = (29 x 19680)/2 = 570720/2 = 285360

Hence, the sum of the series = 285360/99 = 95120/33

Given condition: 

Numbers between 300 and 1000 are divisible by 7.

Concept:

Arithmetic Progression

a_n = a + (n - 1)d 

Calculation:

The first number that is divisible by 7 (300 - 1000) = 301 

Likewise: 301, 308, 315, 322...........994 

The above series makes an AP,

Where a = 301, Common Difference/d = 308 - 301 = 7 and Last term (an) = 994

⇒ an = a + (n - 1)d

⇒ 994 = 301 + (n - 1)7 

⇒ (994 - 301)/7 = n - 1 

⇒ 693/7 + 1 = n 

⇒ 99 + 1 = n 

⇒ n = 100 

∴ There are 100 numbers between 300 and 1000 which are divisible by 7. 

Given:

The sum of even numbers from 21 to 199 is added to 11 observations whose mean value is n.

The mean of the new set of numbers = 99.

Formula used:

(1) Sum of n numbers in A.P.

S = \(\frac{n(a+l)}{2}\)

Where, 

a, is the value of the first term

l, is the value of the last term

n, is the number of terms

S, is the sum of n numbers in A.P

(2) The value of the last term in A.P.

l = a + (n - 1)d

Where, 

a, is the value of the first term

d, is the common difference between two terms

n, is the number of terms

l, is the value of the last term

Calculation:

Let n be the number of even terms between 21 to 199.

The value of the first even number (between 21 to 199), a = 22

The value of the last even number (between 21 to 199), l = 198

The value of the common difference between two even numbers, d = 2

Now,

⇒ 198 = 22 + (n - 1) × 2

⇒ 198 = 22 + (n - 1)2

⇒ 176 = (n - 1)2

⇒ (n - 1) = 88

⇒ n = 89

Now,

Let S be the sum of all even numbers between 21 to 199.

⇒ S = \(\frac{89(22 + 198)}{2}\)

⇒ S = 9790

Now, 

The average of 11 observations = n

The sum of all 11 observations = 11n

According to the question,

⇒ \(\frac{9790+11n}{89+11}\) = 99

⇒ \(\frac{9790+11n}{100}\) = 99

⇒ 9790 + 11n = 9900

⇒ 11n = 110

⇒ n = 10

∴ The required answer is 10.

Additional InformationFormula is used to find the average of numbers when the first and last term is known.

A = \(\frac{a+l}{2}\)

Where, 

a, is the first term of the Arithmetic Progression

l, is the last term of the Arithmetic Progression

A, is the average of Arithmetic Progression from a to l.

Note: The above formula is only applied for Arithmetic Progression.

If the successive terms have a common difference as a non-zero constant, then that sequence can be termed an Arithmetic sequence. 

Given:

First term 'a' = 5, common difference 'd' = 4

Number of terms 'n' = 20

Concept:

Arithmetic progression:

• 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.
• The fixed number is called common difference 'd'.
• It can be positive, negative or zero.

Formula used:

n^th term of AP 

T_n = a + (n - 1)d

The Sum of n terms of AP is given by

\(S = \dfrac{n}{2}[2a + (n-1)d]\)

\(S = \dfrac{n}{2}( a + l)\)

Where, 

a = first term of AP, d = common difference, l = last term 

Calculation:

We know that sum of n terms of AP is given by

\(S = \dfrac{n}{2}[2a + (n-1)d]\)

\(⇒ S = \dfrac{20}{2}[2× 5 + (20-1)× 4]\)

⇒ S = 10(10 + 76)

⇒ S = 860

Hence, the sum of 20 terms given AP will be 860.

We know that n^th term of AP is given by

Tn = a + (n - 1)d

If l is the 20th term (last term) of AP, then

l = 5 + (20 - 1) × 4 = 81

So the sum of AP

\(S = \dfrac{n}{2}( a + l)\)

\(⇒ S = \dfrac{20}{2}(5 + 81)\)

⇒ S = 860

Given

2, 7, 12, ____________

Concept used

Tn = a + (n - 1)d

Where a = first term, n = number of terms and d = difference

Calculation

in the given series

a = 2

d = 7 - 2 = 5

T₁₀ = 2 + (10 - 1) 5

T₁₀ = 2 + 45

T₁₀ = 47

Tenth term = 47

Given: 

For a value of k; 2, 3 + k and 6 are to be in A.P 

Concept:

According to Arithmetic progression, a2 - a1 = a3 - a2 

where a1 ,a2 ,a3 are 1^st, 2^nd and 3^rd term of any A.P.

Calculation:

a₁ = 2, a₂ = k + 3, a₃ = 6 are three consecutive terms of an A.P.

According to Arithmetic progression, a₂ - a₁ = a₃ - a₂ 

(k + 3) – 2 = 6 – (k + 3)

⇒ k + 3 - 8 + k + 3 = 0

⇒ 2k = 2

After solving, we get k = 1

Given:

An AP is given
3 + 7 + 11 + 15 + 19 + ... upto 80 terms 

Formula used:

Sum of n^th term of an AP

S_n = (n/2){2a + (n - 1)d}

where, 

'n' is Number of terms, 'a' is First term, 'd' is Common difference

Calculations:

According to the question, we have

Sn = (n/2){2a + (n - 1)d}      ----(1) 

where, a = 3, n = 80, d = 7 - 3 = 4

Put these values in (1), we get

⇒ S80 = (80/2){2 × 3 + (80 - 1) × 4}

⇒ S80 = 40(6 + 79 × 4)

⇒ S80 = 40 × 322

⇒ S80 = 12,880

∴ The sum of 80^th terms of an AP is 12,880.

Alternate Method

n^th term = a + (n - 1)d

Here n = 80, a = 3 and d = 4

⇒ 80th term = 3 + (80 - 1)4

⇒ 80th term = 3 + 316

⇒ 80th term = 319

Now, the sum of n^th terms of an AP

⇒ S_n = (n/2) × (1st term + Last term)

⇒ S80 = (80/2) × (3 + 319)

⇒ S80 = 40 × 322

⇒ S80 = 12,880

∴ The sum of 80th terms of an AP is 12,880.