Quantitative Aptitude
Number Series
2,662 Questions
Number Series Questions
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$\displaystyle \frac{1}{8}(n+1)[n^{3}+7n^{2}-3n-1]$
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$\displaystyle \frac{n}{8}(n^{2}+4n^{2}+10n+6)$
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$\displaystyle \frac{1}{8}(n+1)[n^{2}+7n^{2}-3n-1]$
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$\displaystyle \frac{1}{8}(n+1)[n^{2}+7n^{2}-3n]$
A
Correct answer
Explanation
For odd n, the terms consist of cubes of odd numbers and 3 times the squares of even numbers. Summing these two parts and simplifying gives (n + 1)(n^3 + 7n^2 - 3n - 1)/8.
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$n^2$
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$n(n+1)$
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$n\left (n+\dfrac {1}{n}\right)^2$
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None
A
Correct answer
Explanation
The series is an arithmetico-geometric series: 1 + 2r + 3r^2 + ... + nr^(n-1) where r = 1 + 1/n. The sum S = 1 + 2r + 3r^2 + ... + nr^(n-1). Then rS = r + 2r^2 + ... + (n-1)r^(n-1) + nr^n. Subtracting gives (1-r)S = 1 + r + r^2 + ... + r^(n-1) - nr^n = (r^n - 1)/(r-1) - nr^n. With r-1 = 1/n, we get (-1/n)S = ( (1+1/n)^n - 1 ) / (1/n) - n(1+1/n)^n. This simplifies to S = n^2.
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$\displaystyle -\frac{5}{6}$
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$\displaystyle -\frac{1}{2}$
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$1$
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$\displaystyle -\frac{3}{2}$
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$10(2^n-1)+n^2$
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$10(2^n+1)+n^2$
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$20(2^n-n)+n^2$
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None of these
B
Correct answer
Explanation
The series is (1-3) + (4-6) + (7-9) + ... for 12 pairs. Each pair sums to -2. 12 * -2 = -24.
B
Correct answer
Explanation
The n-th term a_n is given by S_n - S_(n-1). For n=3, a_3 = S_3 - S_2. S_3 = 3^2 * (3+1) = 9 * 4 = 36. S_2 = 2^2 * (2+1) = 4 * 3 = 12. Thus, a_3 = 36 - 12 = 24.
B
Correct answer
Explanation
The series is an arithmetic progression with first term a = 8 and common difference d = -2. Using the sum formula S = n/2 * (2a + (n-1)d), we set -52 = n/2 * (16 + (n-1)(-2)). Simplifying leads to -104 = 16n - 2n^2 + 2n, or n^2 - 9n - 52 = 0, which factors to (n-13)(n+4) = 0, giving n = 13.
A
Correct answer
Explanation
The first series is 3, 7, 11... (a=3, d=4). The second series is 1, 6, 11... (a=1, d=5). Common terms follow an arithmetic progression with the first term 11 and common difference LCM(4, 5) = 20. The 10th term is a + (n-1)d = 11 + (10-1)20 = 11 + 180 = 191.
D
Correct answer
Explanation
This is an arithmetic progression with a = 20 and d = -2/3. Using the sum formula S = (n/2)(2a + (n-1)d) = 300, we get a quadratic equation 300 = (n/2)(40 - (2/3)(n-1)). Solving this yields n = 25 and n = 36.
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It is an $AP$ and $d=4$, other terms $6,10,14$
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It is an $AP$ and $d=\dfrac{3}{5}$, other terms $5,10,15$
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It is not an $AP$, other terms $3,4,5$
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None of these
A
Correct answer
Explanation
The series is -10, -6, -2, 2. The difference between consecutive terms is -6 - (-10) = 4, -2 - (-6) = 4, 2 - (-2) = 4. It is an AP with d=4. The next terms are 2+4=6, 6+4=10, 10+4=14.
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$121(\sqrt6+\sqrt2)$
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$ \frac{121}{2}(\sqrt3+1)$
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$243(\sqrt3+1)$
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$243(\sqrt3-1)$
A
Correct answer
Explanation
The series is sqrt(2), sqrt(6), sqrt(18), sqrt(54). This is a geometric series with a = sqrt(2) and r = sqrt(3). Sum = a(r^n - 1) / (r - 1) = sqrt(2)(sqrt(3)^10 - 1) / (sqrt(3) - 1) = sqrt(2)(243 - 1) / (sqrt(3) - 1) = 242 * sqrt(2) / (sqrt(3) - 1). Rationalizing gives 121 * sqrt(2) * (sqrt(3) + 1) = 121(sqrt(6) + sqrt(2)).
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$6\, \times\, 2^{10}\, -\, 1$
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$4\, \times\, 2^{10}\, +\, 1$
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$4(2^{10}\, -\, 1)$
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$4(2^{10}\, +\, 1)$
C
Correct answer
Explanation
This is a geometric progression with a = 4, r = 2, and n = 10. The sum is a(r^n - 1) / (r - 1) = 4(2^10 - 1) / (2 - 1) = 4(2^10 - 1).
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$\displaystyle 3+\frac{p}{n}$
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$\displaystyle 3-\frac{p}{n}$
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$\displaystyle 3+\frac{n}{p}$
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$\displaystyle 3-\frac{n}{p}$
B
Correct answer
Explanation
The series is (3 - 1/n) + (3 - 2/n) + (3 - 3/n) + ... The p-th term is clearly 3 - p/n.
B
Correct answer
Explanation
This is an arithmetic progression with a = 31/2, d = 13 - 15.5 = -2.5. Last term = -47. -47 = 15.5 + (n-1)(-2.5). -62.5 = (n-1)(-2.5). n-1 = 25. n = 26.
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$10^{th}$
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$11^{th}$
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$12^{th}$
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$13^{th}$
C
Correct answer
Explanation
This is an arithmetic progression with a = 20 and d = -2. The n-th term is a + (n-1)d. -2 = 20 + (n-1)(-2) => -22 = (n-1)(-2) => 11 = n-1 => n = 12.