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Sequences and Series — JEE Main Mathematics MCQs with Solutions
Free JEE Main Mathematics Sequences and Series MCQs with step-by-step solutions covering Basic Definitions, Arithmetic Progression (A.P.), Geometric Progression (G.P.), Relationship Between A.M. and G.M., Harmonic Progression (H.P.), A.M. – G.M. – H.M. Inequality. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — Basic Definitions · easy · theory
A sequence is best described as a function whose domain is:
A. the empty set
B. the set of rational numbers
C. the set of real numbers
D. the set of natural numbers ✓ Correct
Solution: A sequence is a function $f:\mathbb{N} \to \mathbb{R}$; its values $f(1), f(2), \dots$ are the terms.
Q2 — Basic Definitions · easy · theory
A series is obtained from a sequence by:
A. multiplying its terms
B. reversing its terms
C. adding its terms ✓ Correct
D. taking reciprocals of its terms
Solution: If $a_1, a_2, \dots$ is a sequence, the expression $a_1 + a_2 + \dots$ is the corresponding series.
Q3 — Basic Definitions · easy · numerical
If the general term of a sequence is $T_n = 2n + 3$, then $T_5$ equals:
A. $11$
B. $10$
C. $15$
D. $13$ ✓ Correct
Solution: $T_5 = 2(5) + 3 = 13$.
Q4 — Basic Definitions · easy · numerical
For the sequence with $T_n = n^2 - n$, the value of $T_4$ is:
A. $8$
B. $16$
C. $12$ ✓ Correct
D. $20$
Solution: $T_4 = 4^2 - 4 = 16 - 4 = 12$.
Q5 — Basic Definitions · easy · numerical
The general term of the sequence $\tfrac12, \tfrac23, \tfrac34, \tfrac45, \dots$ is:
A. $\dfrac{n}{n-1}$
B. $\dfrac{n}{n+1}$ ✓ Correct
C. $\dfrac{n+1}{n}$
D. $\dfrac{1}{n+1}$
Solution: Each term is $\tfrac{n}{n+1}$ for $n = 1, 2, 3, \dots$
Q6 — Arithmetic Progression (A.P.) · easy · numerical
The $10$-th term of the A.P. $3, 7, 11, \dots$ is:
A. $37$
B. $43$
C. $39$ ✓ Correct
D. $40$
Solution: $T_{10} = a + 9d = 3 + 9(4) = 39$.
Q7 — A.M. – G.M. – H.M. Inequality · easy · theory
For two distinct positive numbers, the correct ordering of the means is:
A. $A.M. > G.M. > H.M.$ ✓ Correct
B. $G.M. > A.M. > H.M.$
C. $A.M. = G.M. = H.M.$
D. $H.M. > G.M. > A.M.$
Solution: For distinct positive numbers, $A > G > H$; equality holds only when the numbers are equal.
Q8 — A.M. – G.M. – H.M. Inequality · easy · theory
For two positive numbers, the three means $A, G, H$ always satisfy:
A. $G^2 = A \cdot H$ ✓ Correct
B. $A = G = H$
C. $H^2 = A \cdot G$
D. $A^2 = G \cdot H$
Solution: The G.M. is the geometric mean of the A.M. and H.M.: $G^2 = AH$, so $A, G, H$ are in G.P.
Q9 — Summation of Special Series · easy · numerical
The sum of the first $100$ natural numbers $\sum_{r=1}^{100} r$ is:
A. $10100$
B. $5050$ ✓ Correct
C. $4950$
D. $5000$
Solution: $\dfrac{n(n+1)}{2} = \dfrac{100 \cdot 101}{2} = 5050$.
Q10 — Summation of Special Series · easy · numerical
The sum $2 + 4 + 6 + \cdots + 2n$ for $n = 25$ is:
A. $650$ ✓ Correct
B. $625$
C. $675$
D. $600$
Solution: $= n(n+1) = 25 \cdot 26 = 650$.
Q11 — Miscellaneous Series · easy · numerical
The recurring decimal $0.\overline{3}$ (i.e. $0.333\dots$) expressed as a fraction is:
A. $\dfrac13$ ✓ Correct
B. $\dfrac{33}{100}$
C. $\dfrac{3}{10}$
D. $\dfrac{1}{30}$
Solution: As an infinite G.P. $\dfrac{3}{10} + \dfrac{3}{100} + \cdots = \dfrac{3/10}{1 - 1/10} = \dfrac13$.
Q12 — Miscellaneous Series · easy · numerical
The value of $0.\overline{6}$ as a fraction is:
A. $\dfrac35$
B. $\dfrac67$
C. $\dfrac23$ ✓ Correct
D. $\dfrac{6}{11}$
Solution: $\dfrac{6/10}{1 - 1/10} = \dfrac{6}{9} = \dfrac23$.
Q13 — Miscellaneous Series · easy · numerical
The value of $0.\overline{7}$ as a fraction is:
A. $\dfrac{70}{99}$
B. $\dfrac{7}{11}$
C. $\dfrac79$ ✓ Correct
D. $\dfrac{7}{10}$
Solution: A single repeating digit over $9$: $0.\overline{7} = \dfrac79$.
Q14 — Geometric Progression (G.P.) · hard · numerical
If the sum of the first $n$ terms of the G.P. $5 + 10 + 20 + \cdots$ is $635$, then $n$ equals:
A. $9$
B. $6$
C. $7$ ✓ Correct
D. $8$
Solution: $\dfrac{5(2^n - 1)}{2-1} = 635 \Rightarrow 2^n - 1 = 127 \Rightarrow 2^n = 128 \Rightarrow n = 7$.
Q15 — Geometric Progression (G.P.) · hard · numerical
The geometric mean of the five numbers $2, 4, 8, 16, 32$ is:
A. $10$
B. $12$
C. $16$
D. $8$ ✓ Correct
Solution: G.M. $= (2 \cdot 4 \cdot 8 \cdot 16 \cdot 32)^{1/5} = (2^{15})^{1/5} = 2^3 = 8$.
Q16 — Geometric Progression (G.P.) · hard · numerical
The sum to infinity of a G.P. is $15$ and the sum of the squares of its terms is $45$. Its first term is:
A. $5$ ✓ Correct
B. $3$
C. $9$
D. $6$
Solution: $\dfrac{a}{1-r} = 15$ and $\dfrac{a^2}{1-r^2} = 45$; dividing gives $\dfrac{a}{1+r} = 3$, so $r = \tfrac23, a = 5$.
Q17 — Geometric Progression (G.P.) · hard · numerical
The value of $9^{1/3} \cdot 9^{1/9} \cdot 9^{1/27} \cdots$ (to infinity) is:
A. $3$ ✓ Correct
B. $9$
C. $27$
D. $\sqrt{3}$
Solution: Exponent $= \tfrac13 + \tfrac19 + \cdots = \dfrac{1/3}{1 - 1/3} = \tfrac12$, so the product is $9^{1/2} = 3$.
Q18 — Geometric Progression (G.P.) · hard · numerical
In a G.P., the sum of the first two terms is $12$ and the sum of the next two terms is $48$. The common ratio (positive) is:
A. $\tfrac12$
B. $3$
C. $2$ ✓ Correct
D. $4$
Solution: $(ar^2 + ar^3) = r^2(a + ar) \Rightarrow 48 = 12r^2 \Rightarrow r^2 = 4 \Rightarrow r = 2$.
Q19 — Geometric Progression (G.P.) · hard · numerical
For a G.P. with first term $a$ and ratio $r$ ($|r|<1$), if $S_\infty = 4$ and $a = 3$, then $r$ is:
A. $\tfrac13$
B. $\tfrac34$
C. $\tfrac14$ ✓ Correct
D. $\tfrac12$
Solution: $\dfrac{3}{1-r} = 4 \Rightarrow 1 - r = \tfrac34 \Rightarrow r = \tfrac14$.
Q20 — Geometric Progression (G.P.) · hard · numerical
If $a, b, c, d$ are in G.P., then $(b - c)^2 + (c - a)^2 + (d - b)^2$ equals:
A. $(b - c)^2$
B. $(a + d)^2$
C. $0$
D. $(a - d)^2$ ✓ Correct
Solution: A standard G.P. identity: $(b-c)^2 + (c-a)^2 + (d-b)^2 = (a-d)^2$.
Q21 — Relationship Between A.M. and G.M. · hard · numerical
For positive $a, b, c$, the minimum value of $(a+b+c)\left(\dfrac1a + \dfrac1b + \dfrac1c\right)$ is:
A. $3$
B. $6$
C. $1$
D. $9$ ✓ Correct
Solution: By A.M.–G.M. on each factor, the product $\ge 9$, with equality when $a = b = c$.
Q22 — Relationship Between A.M. and G.M. · hard · numerical
The minimum value of $\sin^2\theta + \csc^2\theta$ (where $\sin\theta \ne 0$) is:
A. $2$ ✓ Correct
B. $0$
C. $\tfrac12$
D. $1$
Solution: With $t = \sin^2\theta$, $t + \tfrac1t \ge 2$; the minimum $2$ occurs at $\sin^2\theta = 1$.
Q23 — Relationship Between A.M. and G.M. · hard · numerical
If $x, y, z > 0$ and $x + y + z = 12$, the maximum value of $xyz$ is:
A. $64$ ✓ Correct
B. $81$
C. $27$
D. $48$
Solution: By A.M.–G.M., $xyz \le \left(\tfrac{x+y+z}{3}\right)^3 = 4^3 = 64$, attained at $x = y = z = 4$.
Q24 — Relationship Between A.M. and G.M. · hard · numerical
For $x > 0$, the maximum value of $\dfrac{x}{1 + x^2}$ is:
A. $\tfrac12$ ✓ Correct
B. $\tfrac14$
C. $2$
D. $1$
Solution: Write it as $\dfrac{1}{x + \tfrac1x}$; since $x + \tfrac1x \ge 2$, the maximum is $\tfrac12$ at $x = 1$.
Q25 — Relationship Between A.M. and G.M. · hard · numerical
For positive reals $a, b$, if $A.M. - G.M. = \dfrac{(\sqrt{a} - \sqrt{b})^2}{2}$, then $A.M. = G.M.$ exactly when:
A. $a = b$ ✓ Correct
B. $a = 2b$
C. $a + b = 0$
D. $ab = 1$
Solution: The difference $\tfrac{(\sqrt a - \sqrt b)^2}{2}$ is zero iff $\sqrt a = \sqrt b$, i.e. $a = b$.
Q26 — Arithmetic Progression (A.P.) · hard · numerical
The sum of all two-digit numbers divisible by $3$ is:
A. $1665$ ✓ Correct
B. $1550$
C. $1683$
D. $1584$
Solution: They are $12, 15, \dots, 99$: $30$ terms, sum $= \tfrac{30}{2}(12+99) = 15 \times 111 = 1665$.
Q27 — Harmonic Progression (H.P.) · hard · numerical
If the harmonic mean of two numbers is $4$ and their arithmetic mean is $5$, their geometric mean is:
A. $2\sqrt{5}$ ✓ Correct
B. $4.5$
C. $4$
D. $3$
Solution: Using $G^2 = A \cdot H = 5 \times 4 = 20$, we get $G = 2\sqrt{5}$.
Q28 — Harmonic Progression (H.P.) · hard · numerical
In an H.P., the $5$-th term is $\tfrac{1}{16}$ and the $8$-th term is $\tfrac{1}{25}$. Its general ($n$-th) term is:
A. $\dfrac{1}{4n}$
B. $\dfrac{1}{2n+3}$
C. $\dfrac{1}{3n-1}$
D. $\dfrac{1}{3n+1}$ ✓ Correct
Solution: Reciprocals: $a + 4d = 16, a + 7d = 25 \Rightarrow d = 3, a = 4$, so the $n$-th reciprocal is $3n + 1$.
Q29 — Harmonic Progression (H.P.) · hard · numerical
If $a, b, c$ are in H.P. with $a = 4$ and $c = 12$, then $b$ is:
A. $9$
B. $8$
C. $6$ ✓ Correct
D. $5$
Solution: $b = \dfrac{2 \cdot 4 \cdot 12}{4 + 12} = \dfrac{96}{16} = 6$.
Q30 — Harmonic Progression (H.P.) · hard · theory
If $A$, $G$, $H$ are the A.M., G.M. and H.M. of two positive numbers, then $A$, $G$, $H$ are in:
A. G.P. (i.e. $G^2 = AH$) ✓ Correct
B. H.P.
C. no fixed relation
D. A.P.
Solution: It is a standard result that $G^2 = A \cdot H$, so $A, G, H$ are in G.P.