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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 — Geometric Progression (G.P.) · easy · numerical
The $6$-th term of the G.P. $3, 6, 12, \dots$ is:
A. $96$ ✓ Correct
B. $192$
C. $48$
D. $64$
Solution: $T_6 = a r^5 = 3 \cdot 2^5 = 96$.
Q8 — Relationship Between A.M. and G.M. · easy · theory
For two positive real numbers, the relation between their A.M. and G.M. is:
A. $A.M. < G.M.$
B. $A.M. \le G.M.$
C. $A.M. = G.M.$ always
D. $A.M. \ge G.M.$ ✓ Correct
Solution: For positive reals, $\dfrac{a+b}{2} \ge \sqrt{ab}$, with equality iff $a = b$.
Q9 — Relationship Between A.M. and G.M. · easy · numerical
The A.M. and G.M. of $4$ and $9$ are respectively:
A. $5$ and $6$
B. $6.5$ and $6$ ✓ Correct
C. $6$ and $6.5$
D. $6.5$ and $6.5$
Solution: A.M. $= \tfrac{13}{2} = 6.5$, G.M. $= \sqrt{36} = 6$; indeed A.M. $>$ G.M.
Q10 — Harmonic Progression (H.P.) · easy · theory
A sequence is a Harmonic Progression (H.P.) if:
A. the reciprocals of its terms form a G.P.
B. the reciprocals of its terms form an A.P. ✓ Correct
C. its terms form a G.P.
D. its terms form an A.P.
Solution: By definition, $a_1, a_2, \dots$ is an H.P. iff $\tfrac{1}{a_1}, \tfrac{1}{a_2}, \dots$ is an A.P.
Q11 — Harmonic Progression (H.P.) · easy · theory
The reciprocals of the H.P. $1, \tfrac13, \tfrac15, \tfrac17, \dots$ form the A.P.:
A. $1, 2, 3, 4, \dots$
B. $2, 4, 6, \dots$
C. $1, 3, 5, 7, \dots$ ✓ Correct
D. $1, \tfrac13, \tfrac15, \dots$
Solution: Reciprocals are $1, 3, 5, 7, \dots$, an A.P. with common difference $2$.
Q12 — Harmonic Progression (H.P.) · easy · numerical
The harmonic mean of $4$ and $6$ is:
A. $4.8$ ✓ Correct
B. $5.2$
C. $5$
D. $4.5$
Solution: H.M. $= \dfrac{2ab}{a+b} = \dfrac{2 \cdot 24}{10} = 4.8$.
Q13 — Harmonic Progression (H.P.) · easy · numerical
The harmonic mean of $2$ and $8$ is:
A. $3.2$ ✓ Correct
B. $5$
C. $4$
D. $3.5$
Solution: H.M. $= \dfrac{2 \cdot 16}{10} = 3.2$.
Q14 — 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.
Q15 — 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.
Q16 — 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: $\dfrac{n(n+1)}{2} = \dfrac{100 \cdot 101}{2} = 5050$.
Q17 — 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: $= n(n+1) = 25 \cdot 26 = 650$.
Q18 — 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. $\dfrac{3}{10} + \dfrac{3}{100} + \cdots = \dfrac{3/10}{1 - 1/10} = \dfrac13$.
Q19 — 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: $\dfrac{6/10}{1 - 1/10} = \dfrac{6}{9} = \dfrac23$.
Q20 — 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 $9$: $0.\overline{7} = \dfrac79$.
Q21 — Basic Definitions · hard · numerical
If the sum of $n$ terms of a sequence is $S_n = 2^n - 1$, then its $10$-th term is:
A. $1023$
B. $256$
C. $1024$
D. $512$ ✓ Correct
Solution: $T_{10} = S_{10} - S_9 = (2^{10}-1) - (2^9 - 1) = 1024 - 512 = 512 = 2^9$.
Q22 — Basic Definitions · hard · numerical
The general term of the sequence $5, 11, 19, 29, \dots$ (successive differences $6, 8, 10, \dots$) is:
A. $3n^2 - 1$
B. $n^2 + 3n + 1$ ✓ Correct
C. $2n^2 + 3$
D. $n^2 + 2n + 2$
Solution: Differences form an A.P., so $T_n$ is quadratic: $T_n = n^2 + 3n + 1$ fits $5, 11, 19, 29$.
Q23 — Basic Definitions · hard · numerical
If $T_n = \dfrac{1}{n(n+1)}$, then $T_1 + T_2 + T_3$ equals:
A. $\dfrac{2}{3}$
B. $\dfrac{11}{12}$
C. $\dfrac{3}{4}$ ✓ Correct
D. $\dfrac{1}{4}$
Solution: $\tfrac12 + \tfrac16 + \tfrac{1}{12} = \tfrac{6+2+1}{12} = \tfrac{9}{12} = \tfrac34$.
Q24 — Basic Definitions · hard · numerical
The $n$-th term of the sequence $2, 5, 10, 17, \dots$ increased by the $n$-th term of $1, 2, 3, 4, \dots$ gives, for $n = 4$:
A. $21$ ✓ Correct
B. $18$
C. $20$
D. $26$
Solution: First sequence $T_n = n^2 + 1 \Rightarrow T_4 = 17$; adding $4$ gives $21$.
Q25 — Arithmetic Progression (A.P.) · hard · numerical
If the $8$-th (middle) term of an A.P. is $20$, the sum of its first $15$ terms is:
A. $320$
B. $150$
C. $280$
D. $300$ ✓ Correct
Solution: For an odd number of terms, $S_{2k-1} = (2k-1)\,a_k$; here $S_{15} = 15 \times 20 = 300$.
Q26 — Arithmetic Progression (A.P.) · hard · numerical
If the $p$-th term of an A.P. is $q$ and the $q$-th term is $p$ (with $p \ne q$), then the $(p+q)$-th term is:
A. $p + q$
B. $1$
C. $0$ ✓ Correct
D. $pq$
Solution: From $a+(p-1)d=q$ and $a+(q-1)d=p$, we get $d=-1$ and $a=p+q-1$; hence $T_{p+q}=a+(p+q-1)d=0$.
Q27 — Arithmetic Progression (A.P.) · hard · numerical
The sum of $n$ terms of an A.P. is $3n^2 + 5n$. If its $m$-th term is $164$, then $m$ equals:
A. $26$
B. $25$
C. $27$ ✓ Correct
D. $28$
Solution: $T_n = S_n - S_{n-1} = 6n - 1$? Compute: $6n + 2$. Then $6m + 2 = 164 \Rightarrow m = 27$.
Q28 — Arithmetic Progression (A.P.) · hard · numerical
The angles of a triangle are in A.P. and the greatest is twice the least. The angles are:
A. $30^\circ, 60^\circ, 90^\circ$
B. $40^\circ, 60^\circ, 80^\circ$ ✓ Correct
C. $50^\circ, 60^\circ, 70^\circ$
D. $45^\circ, 60^\circ, 75^\circ$
Solution: Middle angle $= 60^\circ$; with $60+d = 2(60-d)$ we get $d = 20$, giving $40, 60, 80$.
Q29 — Arithmetic Progression (A.P.) · hard · numerical
If the $18$-th and $11$-th terms of an A.P. are in the ratio $3:2$, the ratio of its $21$-st and $5$-th terms is:
A. $2 : 1$
B. $5 : 2$
C. $3 : 2$
D. $3 : 1$ ✓ Correct
Solution: $\tfrac{a+17d}{a+10d} = \tfrac32 \Rightarrow a = 4d$; then $\tfrac{a+20d}{a+4d} = \tfrac{24d}{8d} = 3:1$.
Q30 — Arithmetic Progression (A.P.) · hard · numerical
The sum of the first $n$ terms of two A.P.s are in the ratio $(7n+1):(4n+27)$. The ratio of their $11$-th terms is:
A. $7 : 4$
B. $11 : 27$
C. $148 : 111$ ✓ Correct
D. $4 : 7$
Solution: Ratio of $m$-th terms $= \tfrac{7(2m-1)+1}{4(2m-1)+27}$; for $m=11$, $2m-1 = 21$: $\tfrac{148}{111}$.