Quantitative Aptitude
Number Series
2,662 Questions
Number Series Questions
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$\dfrac{625}{3}$
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$\dfrac{125}{3}$
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$81$
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$9$
D
Correct answer
Explanation
The sequence is 2^1, 2^2, 2^3, 2^4. The next term is 2^5 = 32.
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$0.00001$
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$0.0001$
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$0.000001$
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$0.00000001$
A
Correct answer
Explanation
The sequence is 10^-1, 10^-2, 10^-3, 10^-4. The next term is 10^-5 = 0.00001.
A
Correct answer
Explanation
The sequence is a geometric progression with common ratio -1/2. -24 * (-1/2) = 12.
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$7$
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$5$
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$3$
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$None\ of\ these$
A
Correct answer
Explanation
This is a geometric series with a = 1 and r = 3. Sum = a(r^n - 1) / (r - 1) = (3^n - 1) / 2. We want (3^n - 1) / 2 > 1000, so 3^n - 1 > 2000, 3^n > 2001. 3^6 = 729, 3^7 = 2187. Thus, n must be at least 7.
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$\displaystyle\frac{8.10^{10}+1}{27.10^9}$
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$\displaystyle\frac{9.10^{10}+1}{27.10^9}$
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$\displaystyle\frac{9.10^9+1}{27.10^8}$
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$\displaystyle\frac{8.10^9+1}{27.10^8}$
A
Correct answer
Explanation
The series is 0.33 + 0.333 + ... = (1/3)(0.99 + 0.999 + ...) = (1/3)((1-0.1) + (1-0.01) + ...). Sum = (1/3)(n - sum of geometric series). For n=9, this simplifies to the given expression.
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$\cfrac { 1 }{ 2 } \left( { 3 }^{ n }+1 \right) $
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$\cfrac { 1 }{ 2 } \left( { 3 }^{ n }-1 \right) $
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$\left( \cfrac { { 2 }^{ n }+1 }{ 2 } \right) 2$
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${ 3 }^{ n }-1$
B
Correct answer
Explanation
The series is 1, 4, 13, 40, 121, 364. This is a geometric series sum: 1 + 3 + 9 + 27 + 81 + 243. The nth term is the sum of a geometric progression with first term 1 and ratio 3: (3^n - 1) / (3 - 1) = (3^n - 1) / 2.
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$n$th term $T_{n}=(2^{n}-1)$. Sum of $n$ terms$=2^{n}-1-n$
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$n$th term $T_{n}=(2^{n}-1)$. Sum of $n$ terms$=2(2^{n}-1)-n$
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$n$th term $T_{n}=(2^{n}-1)$. Sum of $n$ terms$=2(2^{n}-1)-2n$
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$n$th term $T_{n}=(2^{n}-1-n)$. Sum of $n$ terms$=2(2^{n}-1)-n-1$
B
Correct answer
Explanation
The series is 1, 3, 7, 15, 31... which is 2^1-1, 2^2-1, 2^3-1, 2^4-1, 2^5-1. Thus Tn = 2^n - 1. The sum is sum(2^k - 1) from k=1 to n = (2^(n+1) - 2) - n = 2(2^n - 1) - n.
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$\displaystyle \frac{(10^{n}-1)}{9}+2^{n}$
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$\displaystyle \frac{10(10^{n}-1)}{9}+2(2^{n}-1)$
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$\displaystyle \frac{10^{n}-1}{9}-1$
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$\displaystyle \frac{10(10^{n}-1)}{9}$
B
Correct answer
Explanation
The series is (10+2) + (100+4) + (1000+8) + ... = (10^1 + 10^2 + ... + 10^n) + (2^1 + 2^2 + ... + 2^n). The first part is a GP sum: 10(10^n - 1) / 9. The second part is a GP sum: 2(2^n - 1) / (2-1) = 2(2^n - 1).
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$\displaystyle { 2 }^{ n }-n-1$
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$\displaystyle 1-{ 2 }^{ -n }$
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$\displaystyle { 2 }^{ -n }+n-1$
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None of these
C
Correct answer
Explanation
The series is 1/2 + 3/4 + 7/8 + ... + (2^n - 1)/2^n. This can be written as (1 - 1/2) + (1 - 1/4) + ... + (1 - 1/2^n) = n - (1/2 + 1/4 + ... + 1/2^n). The sum of the geometric series is (1/2)(1 - (1/2)^n) / (1 - 1/2) = 1 - 2^-n. So the total sum is n - (1 - 2^-n) = n - 1 + 2^-n.
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$\displaystyle \frac { 1 }{ 3 } $
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1
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$\displaystyle \frac { 2 }{ 5 } $
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$\displaystyle \sqrt { \frac { 2 }{ 3 } } $
C
Correct answer
Explanation
The terms are 10/9, (1/3)*sqrt(20/3), 2/3. Simplifying: 10/9, (sqrt(20)/3*sqrt(3)) = (2*sqrt(5))/(3*sqrt(3)), 2/3. This is a geometric progression with first term a = 10/9 and common ratio r = (sqrt(20/3)) / (10/3) = sqrt(20/3) * 3/10 = sqrt(20/3) * sqrt(9/100) = sqrt(20*9 / 3*100) = sqrt(180/300) = sqrt(3/5). The 5th term is a*r^4 = (10/9) * (3/5)^2 = (10/9) * (9/25) = 10/25 = 2/5.
D
Correct answer
Explanation
This is a geometric series with a = -4 and r = -4. The sum formula is S_n = a(r^n - 1) / (r - 1). Substituting: 52428 = -4((-4)^n - 1) / (-5). 52428 * 5 / 4 = (-4)^n - 1. 65535 = (-4)^n - 1. 65536 = (-4)^n. Since (-4)^8 = 65536, n = 8.
C
Correct answer
Explanation
Geometric sequence: a * r^(n-1) = term. a * (-3)^7 = 4374. a * (-2187) = 4374. a = -2.
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$\dfrac {5}{9} \left [n - \dfrac {2}{9} \left (1 - \dfrac {1}{10^{n}}\right )\right ]$
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$\dfrac {1}{9} \left [5 - \dfrac {2}{9} \left (1 - \dfrac {1}{10^{n}}\right )\right ]$
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$\dfrac {1}{9} \left [n - \dfrac {5}{9} \left (1 - \dfrac {1}{10^{n}}\right )\right ]$
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$\dfrac {5}{9} \left [n - \dfrac {1}{9} \left (1 - \dfrac {1}{10^{n}}\right )\right ]$
D
Correct answer
Explanation
The series is 0.5 + 0.55 + 0.555... = 5/9 * (0.9 + 0.99 + 0.999...) = 5/9 * [(1-0.1) + (1-0.01) + (1-0.001)...] = 5/9 * [n - (0.1 + 0.01 + ...)] = 5/9 * [n - 1/9 * (1 - 1/10^n)].