Official curriculum roadmap, subject/topic distribution, negative marking rules, pacing guidelines, and solved sample questions.
๐ฏ Mapped Subjects & Topic Question Distribution
Total Question Pool100%
10 MCQs
Combined Active Syllabus
Sequences
10 MCQs
Topic Pool
๐ Question Pool Structure
10 MCQs across fundamental, intermediate, and advanced concept tiers.
โก Recommended Pacing
45 to 60 seconds per MCQ. Flag complex problems and preserve 10 minutes for final revision.
โ๏ธ Scoring & Negative Marking
+1 mark per correct answer. In competitive tests with negative marking, -0.25 applies for incorrect guesses.
๐ก Strategic Preparation & Exam Hall Guidelines
To maximize your score on Sequences, candidates are advised to follow a structured three-pass approach. In the First Pass, solve all direct recall and formula-based questions within 30 seconds each to secure foundational marks. In the Second Pass, tackle multi-step analytical and quantitative reasoning problems. In the Third Pass, review marked questions and verify calculations.
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Sample Question 1
SequencesMedium • Quantitative Aptitude Test
Find the sum of the first 10 terms of the arithmetic sequence: , 7, 12, 17, \dots$
The first term is = 2$, common difference = 7 - 2 = 5$, and number of terms = 10$. Using the sum formula for an arithmetic progression: \[S_n = \frac{n}{2}[2a + (n-1)d]\] \[S_{10} = \frac{10}{2}[2(2) + (10-1)5] = 5[4 + 45] = 5 \times 49 = 245\] Thus, the correct option is (B).
Sample Question 2
SequencesHard • Quantitative Aptitude Test
Find the 100th term of the arithmetic sequence: $15, 20, 25, 30, \dots$
First term $a = 15$ and common difference $d = 20 - 15 = 5$. The $n$-th term is $a_n = a + (n-1)d$. For $n = 100$: \[a_{100} = 15 + (100 - 1) \times 5 = 15 + 99 \times 5 = 15 + 495 = 510\] Thus, the correct option is (B).
Sample Question 3
SequencesMedium • Quantitative Aptitude Test
Find the sum of the first $50$ odd positive integers: $1 + 3 + 5 + 7 + \dots$
The sum of the first $n$ odd positive integers is given by $S_n = n^2$. For $n = 50$: \[S_{50} = 50^2 = 2500\] Or using arithmetic series formula: $S_{50} = \frac{50}{2}[2(1) + (50-1)2] = 25[2 + 98] = 25 \times 100 = 2500$. Thus, the correct option is (C).
Sample Question 4
SequencesHard • Quantitative Aptitude Test
Find the 8th term in the geometric progression: $1, 3, 9, \dots$
Here first term $a = 6$ and common ratio $r = \frac{12}{6} = 2$. The $n$-th term is: \[a_n = a r^{n-1} \implies 1536 = 6 \times 2^{n-1} \implies 2^{n-1} = \frac{1536}{6} = 256\] Since $256 = 2^8$, we have $n - 1 = 8 \implies n = 9$. Thus, the correct option is (C).
Sample Question 6
SequencesHard • Quantitative Aptitude Test
Find the sum of the infinite geometric progression: $\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots$
Here $a = \frac{1}{3}$ and common ratio $r = \frac{1}{3}$. Using $S_\infty = \frac{a}{1 - r}$: \[S_\infty = \frac{1/3}{1 - 1/3} = \frac{1/3}{2/3} = \frac{1}{2}\] Thus, the correct option is (B).
Sample Question 7
SequencesHard • Quantitative Aptitude Test
In the geometric sequence $1, 3, 9, 27, \dots$, find the common ratio ($r$).
The common ratio $r$ is the quotient of any term divided by the preceding term: \[r = \frac{3}{1} = \frac{9}{3} = \frac{27}{9} = 3\] Thus, the correct option is (B).
Sample Question 8
SequencesHard • Quantitative Aptitude Test
Find the value of the 10th term in the geometric sequence: $1, 3, 9, \dots$
Here the first term is $a_1 = 1$ and common ratio is $r = \frac{3}{1} = 3$. The $n$-th term is $a_n = a_1 r^{n-1}$. For $n = 10$: \[a_{10} = 1 \times 3^{10-1} = 3^9\] Thus, the correct option is (B).
Sample Question 9
SequencesMedium • Quantitative Aptitude Test
If the first term of a geometric sequence is $a_1 = 6$ and the common ratio is $r = 2$, find the 9th term.
The formula for the $n$-th term of a geometric progression is $a_n = a_1 r^{n-1}$. For $n = 9$: \[a_9 = 6 \times 2^{9-1} = 6 \times 2^8 = 6 \times 256 = 1536\] Thus, the correct option is (C).
Sample Question 10
SequencesHard • Quantitative Aptitude Test
Calculate the sum of the infinite geometric series: $\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots$
The first term is $a = \frac{1}{3}$ and common ratio is $r = \frac{1/9}{1/3} = \frac{1}{3}$. Using the infinite sum formula $S_\infty = \frac{a}{1 - r}$: \[S_\infty = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}\] Thus, the correct option is (B).
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