COMEDK UGET 2025 Sequence and Series Practice Questions with Solutions

COMEDK UGET 2025 Sequence and Series Practice Questions with Solutions: Sequences and Series contributes significantly to the Algebra section of COMEDK exam syllabus. The chapter features concepts of Arithmetic Progression and Geometric Progression. If you are planning to appear in COMEDK UGET 2025 exam, you should practice Sequence and Series questions to better understand the nature and difficulty level of the questions, identify the most important and high-weightage topics etc. In this article, we have highlighted Sequence and Series important topics, weightage, and also provided a list of COMEDK UGET 2025 Sequence and Series practice questions with solutions. 

Also Check - Do or Die Chapters for COMEDK UGET 2025 Mathematics

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COMEDK UGET 2025 Sequence and Series Important Topics

Sequence and Series is a part of the Algebra section in COMEDK UGET 2025 Mathematics syllabus. In the following table, we have provided all the subtopics included in Sequence and Series chapter. COMEDK Sequence and Series sample questions are prepared based on the same topics, so you must prepare them well before solving the practice papers. 

COMEDK UGET 2025 Sequence and Series Expected Weightage

As per the past years' question paper trends, you can expect around 10 questions from the COMEDK Sequence and Series chapter. This means, Sequence and Series in COMEDK UGET question paper 2025 will have an approximate weightage of 20%. In the following table, we have listed the expected weightage for COMEDK UGET 2025 Sequence and Series:

COMEDK UGET 2025 Sequence and Series Practice Questions with Solutions

After analyzing the recent question paper trends, our experts have curated some sample questions from the Sequence and Series chapter. You are advised to practice these questions thoroughly while doing your preparation/ revision. You can check COMEDK UGET 2025 Sequence and Series practice questions with solutions below: 

Q1. Consider an infinite geometric series with first term a and common ratio r. If the sum of an infinite geometric series is 4 and the second term is 3/4 then:

a. a = 1, r = -3/4 

b. a = 3, r = 1/4 

c. a = -3, r = -1/4 

d. a = -1, r = 3/4 

Ans. b. a = 3, r = 1/4

Solution: To determine the correct values for a and r, we need to use the properties of infinite geometric series.

The sum of an infinite geometric series can be expressed as:

S = a/1-r

We are given that the sum of the infinite series is 4

a/1-r = 4 …1

The second term of the series can be calculated as

ar

We are given that the second term is 3/4 

ar = 3/4 …2

From equation (2), we can express a in terms of r

a = 3/4r

Substitute this into equation (1):

(3/4r)/1-r = 4

Simplify and solve for r:

3/4r(1-r) = 4

3 = 16r(1-r)

3 = 16r - 16r²

16r² - 16r + 3 = 0

Solving this quadratic equation for r,

r = -b±√(b² - 4ac)/2a

here , a = 16, b = -16, and c = 3

Solving the quadratic equation,

r = 3/4 or r = 1/4 

Now, using these values for r, we find corresponding a

For r = 3/4 

a =3/4(3/4) = 1

Similarly, for r = 1/4, a = 3

The correct pairs of values are:

a = 1, r = 3/4

a = 3, 4 = 1/4  

Q2. The sum of four numbers in a geometric progression is 60, and the arithmetic mean of the first and the last number is 18. Then the numbers are

a. 10, 8, 16, 26

b. 32, 16, 4, 8

c. 32, 16, 8, 2

d. 4, 8, 16, 32

Ans.d. 4, 8, 16, 32

Solution: Let's consider four numbers in a geometric progression. We can denote them as follows:

a, ar, ar², ar³…

Where a is the first term and r is the common ratio

We are given two conditions:

The sum of the four numbers is 60:

a + ar + ar² + ar³ = a(1 + r + r² + r³) = 60

The arithmetic mean of the first and the last number is 18:

a + ar3/2 = 18 = a(1+r³) = 36

Now, let’s check the options provided.

Option D is: 4, 8, 16, 32.

Verify if these are in a geometric progression:

Here, a = 4.

The ratio from 4 to 8 is 8/4 = 2, from 8 to 16 is 16/8 = 2, and from 16 to 32 is 32/16=2, thus, r = 2.

Check the sum

4 + 8 + 16 + 32 = 60

Check the arithmetic mean of the first and last number:

4+32/2 = 36/2 = 18

Since both conditions are satisfied by Option D, the numbers in the geometric progression are:

4, 8, 16, 32.

Q3. If the 6th term of G.P is -1/32 and 9th term is 1/256 the r is?

a. 2

b. -1/2 

c. 1/2

d. -2

Ans.b. -1/2

Solution: Let's solve the problem step by step.

The general term of a geometric progression (GP) is given by:

Tn = ar^n-1

where:

a is the first term,

r is the common ratio.

According to the problem:

The 6th term is:

T₆ = a r⁵ = -1/32

And the 9th term is: 

T₉ = a r⁸ = 1/256

To eliminate a, divide 9th term by 6th term,

T₉/T₆ = a r⁸/a r⁵ = (1/256)/(-1/32)

Simplify the right side:

r³ = 1/256 x (-32/1) = -1/8

Now solve for r,

r³ = -1/8 = r = -1/2

Q4. If the sum of 12th and 22nd terms of an AP is 100, then the sum of the first 33 terms of an AP is?

a. 1700

b. 1650

c. 3300

d. 3500

Ans.b. 1650

Solution: Here, T₁₂ = a + 11d and T₂₂ = a + 21d

Since, 100 = T₁₂ + T₂₂
100 = a + 11d + a + 21 d 

a + 16d = 50 …1

Now, S33 = 33/2[2a + (33 - 1)d] 

33(a+16d)

From equation 1

33 x 50 = 1650

Q5. If three numbers a, b, c constitute both an A.P and G.P, then

a. a = b = c

b. a = b + c

c. ab = c

d. a = b - c

Ans. a. a = b = c

Solution: To solve this, let's first understand what it means for numbers to form an arithmetic progression (A.P) and a geometric progression (G.P).

Arithmetic Progression (A.P): A sequence of numbers is said to be in arithmetic progression when the difference between any two successive members is a constant. For example, in the sequence a, b, c, where b, c are successive terms after a, they must satisfy:

b - a = c - b

Simplifying, we get:

2b = a + c

Geometric Progression (G.P): A sequence is in geometric progression when each term after the first is multiplied by a constant called the common ratio. In the sequence a, b, c, they must satisfy: b/a = c/b

If a, b, and c are non zero, we can rearrange the equation as,

b² = ac 

Now, we know that a, b, c are in AP and GP. The key to solving this is to see what happens when we apply the conditions of both progressions. From the G.P. condition, b² = ac. From the A.P. condition, 2b = a + c if we substitute a + c = 2b into GP equation,

b² = ac

b² = a (2b - a)

Let’s simplify this:

b² = 2ab - a²

This actually is a quadratic equation in terms of a,

a² - 2ab + b² = 0

which simplifies to:

(a - b)² = 0

Thus, a - b = 0

a = b

If a = b, then substituting this back in a + c = 2b 

a + c = 2b

c = a

Thus, a = b = c, which concludes that all three numbers must be equal in both A.P. and G.P. when they are non-zero and effective.

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