Q. 1
Quantitative Aptitude Test
Difficulty: Medium
(1 Mark)
Find the sum of the first 10 terms of the arithmetic sequence: , 7, 12, 17, \dots$
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Step-by-Step Explanation & Concept Rationale
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).
Q. 2
Quantitative Aptitude Test
Difficulty: Hard
(1 Mark)
Find the 100th term of the arithmetic sequence: $15, 20, 25, 30, \dots$
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Step-by-Step Explanation & Concept Rationale
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).
Q. 3
Quantitative Aptitude Test
Difficulty: Medium
(1 Mark)
Find the sum of the first $50$ odd positive integers: $1 + 3 + 5 + 7 + \dots$
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Step-by-Step Explanation & Concept Rationale
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).
Q. 4
Quantitative Aptitude Test
Difficulty: Hard
(1 Mark)
Find the 8th term in the geometric progression: $1, 3, 9, \dots$
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Step-by-Step Explanation & Concept Rationale
First term $a = 1$, common ratio $r = 3$. The 8th term is: \[a_8 = a r^{8-1} = 1 \times 3^7 = 2187\] Thus, the correct option is (B).
Q. 5
Quantitative Aptitude Test
Difficulty: Hard
(1 Mark)
In the geometric sequence $6, 12, 24, \dots$, what is the position (term number) of the value $1536$?
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Step-by-Step Explanation & Concept Rationale
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).
Q. 6
Quantitative Aptitude Test
Difficulty: Hard
(1 Mark)
Find the sum of the infinite geometric progression: $\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots$
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Step-by-Step Explanation & Concept Rationale
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).
Q. 7
Quantitative Aptitude Test
Difficulty: Hard
(1 Mark)
In the geometric sequence $1, 3, 9, 27, \dots$, find the common ratio ($r$).
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Step-by-Step Explanation & Concept Rationale
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).
Q. 8
Quantitative Aptitude Test
Difficulty: Hard
(1 Mark)
Find the value of the 10th term in the geometric sequence: $1, 3, 9, \dots$
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Step-by-Step Explanation & Concept Rationale
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).
Q. 9
Quantitative Aptitude Test
Difficulty: Medium
(1 Mark)
If the first term of a geometric sequence is $a_1 = 6$ and the common ratio is $r = 2$, find the 9th term.
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Step-by-Step Explanation & Concept Rationale
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).
Q. 10
Quantitative Aptitude Test
Difficulty: Hard
(1 Mark)
Calculate the sum of the infinite geometric series: $\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots$
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Step-by-Step Explanation & Concept Rationale
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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