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Sequences & Series — JEE Main Maths PYQ MCQs with Solutions
Free JEE Main Maths PYQ Sequences & Series MCQs with step-by-step solutions covering Arithmetic Progression, Geometric Progression, Some Special Series. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — Arithmetic Progression · easy · theory
The $n$-th term of an AP is $a_n = 3n + 2$. The sum of the first $n$ terms is:
A. $\frac{n(3n+7)}{2}$ ✓ Correct
B. $\frac{n(3n+5)}{2}$
C. $n(3n+2)$
D. $\frac{3n^2+5n}{2}$
Solution: Given $a_n = 3n + 2$. First term $a_1 = 3(1) + 2 = 5$. Second term $a_2 = 3(2) + 2 = 8$. Common difference $d = 8 - 5 = 3$. Sum formula: $S_n = \frac{n}{2}(2a + (n-1)d) = \frac{n}{2}(2(5) + (n-1)3) = \frac{n}{2}(10 + 3n - 3) = \frac{n(3n+7)}{2}$.
Q2 — Arithmetic Progression · easy · theory
In an AP, if $a_1 = 2$, $a_n = 14$, and $S_n = 128$, then $n$ is:
A. 15
B. 16 ✓ Correct
C. 14
D. 17
Solution: Sum formula: $S_n = \frac{n}{2}(a_1 + a_n) = \frac{n}{2}(2 + 14) = 8n$. Given $S_n = 128$, so $8n = 128 \Rightarrow n = 16$.
Q3 — Arithmetic Progression · easy · theory
The arithmetic mean between $a$ and $b$ is:
A. $\frac{a+b}{2}$ ✓ Correct
B. $\sqrt{ab}$
C. $\frac{2ab}{a+b}$
D. $\frac{a-b}{2}$
Solution: The arithmetic mean (AM) between two numbers $a$ and $b$ is the middle term of the AP formed by $a$, AM, $b$. If this middle term is $x$, then $x - a = b - x$ (equal common differences), so $2x = a + b$, hence $x = \frac{a+b}{2}$.
Q4 — Arithmetic Progression · easy · theory
If $a$, $b$, $c$ are in AP, then $b - a$ is equal to:
A. $c - b$ ✓ Correct
B. $c - a$
C. $a + c - 2b$
D. $2b - a - c$
Solution: If $a$, $b$, $c$ are in AP, then $b - a = d$ (common difference) and $c - b = d$. Therefore $b - a = c - b$.
Q5 — Arithmetic Progression · easy · theory
The sum of the first $n$ odd numbers is:
A. $n^2$ ✓ Correct
B. $n(n+1)/2$
C. $2n-1$
D. $n(2n-1)$
Solution: The first $n$ odd numbers are $1, 3, 5, \ldots, (2n-1)$. This is an AP with $a = 1$, $d = 2$, and $l = 2n-1$. Sum = $\frac{n}{2}(1 + 2n - 1) = \frac{n}{2}(2n) = n^2$.
Q6 — Arithmetic Progression · easy · theory
In an AP, the sum of first $n$ terms is $S_n = 3n^2 - 2n$. The common difference is:
A. 4
B. 6 ✓ Correct
C. 5
D. 3
Solution: From $S_n = 3n^2 - 2n$, we get $a_1 = S_1 = 1$ and $a_2 = S_2 - S_1 = 8 - 1 = 7$. Common difference $d = a_2 - a_1 = 6$.
Q7 — Arithmetic Progression · easy · theory
The $n$-th term of an AP is $5n - 3$. The sum of first 20 terms is:
A. 990 ✓ Correct
B. 980
C. 1000
D. 1010
Solution: Given $a_n = 5n - 3$. First term $a_1 = 2$. 20th term $a_{20} = 97$. Sum of first 20 terms: $S_{20} = \frac{20}{2}(2 + 97) = 10 \times 99 = 990$.
Q8 — Arithmetic Progression · easy · theory
If $a_1, a_2, a_3, \ldots$ form an AP with common difference $d$, then $a_1 - a_2 + a_3 - a_4 + \ldots + a_{2n-1} - a_{2n}$ equals:
A. $-nd$ ✓ Correct
B. $nd$
C. $2nd$
D. $0$
Solution: $(a_1 - a_2) + (a_3 - a_4) + \ldots + (a_{2n-1} - a_{2n}) = (-d) + (-d) + \ldots + (-d) = -nd$ (there are $n$ pairs).
Q9 — Arithmetic Progression · easy · theory
The number of terms in the AP $3, 7, 11, 15, \ldots, 399$ is:
A. 99
B. 100 ✓ Correct
C. 101
D. 102
Solution: First term $a = 3$, common difference $d = 4$, last term $l = 399$. Using $l = a + (n-1)d$: $399 = 3 + 4(n-1) \Rightarrow 396 = 4(n-1) \Rightarrow n = 100$.
Q10 — Arithmetic Progression · easy · theory
The sum $1 + 2 + 3 + \ldots + n$ equals:
A. $\frac{n(n+1)}{2}$ ✓ Correct
B. $\frac{n^2}{2}$
C. $n^2$
D. $\frac{n(n-1)}{2}$
Solution: The series $1, 2, 3, \ldots, n$ is an AP with $a = 1$, $d = 1$, $l = n$. Sum = $\frac{n}{2}(1 + n) = \frac{n(n+1)}{2}$.
Q11 — Arithmetic Progression · easy · theory
In an AP, if $a = 10$, $d = 5$, and $a_n = 95$, then $n$ is:
A. 17
B. 18 ✓ Correct
C. 19
D. 20
Solution: Using $a_n = a + (n-1)d$: $95 = 10 + 5(n-1) \Rightarrow 85 = 5(n-1) \Rightarrow n = 18$.
Q12 — Arithmetic Progression · easy · theory
The $n$-th term of an AP is $T_n = 2n + 1$. Find the 10th term:
A. 21 ✓ Correct
B. 22
C. 20
D. 23
Solution: $T_n = 2n + 1$. So $T_{10} = 2(10) + 1 = 21$.
Q13 — Arithmetic Progression · easy · theory
The sum of squares of first $n$ natural numbers is $\frac{n(n+1)(2n+1)}{6}$. If $S = 1^2 + 2^2 + 3^2 + \ldots + 20^2$, then $S$ equals:
A. 2870 ✓ Correct
B. 2880
C. 2860
D. 2850
Solution: Using the formula $S = \frac{20(21)(41)}{6} = \frac{17220}{6} = 2870$.
Q14 — Geometric Progression · easy · theory
If $a$, $b$, $c$ are in GP, then:
A. $a^2 = bc$
B. $b^2 = ac$ ✓ Correct
C. $c^2 = ab$
D. $ab = bc$
Solution: If $a$, $b$, $c$ are in GP with common ratio $r$, then $b = ar$ and $c = ar^2$. Thus $b^2 = (ar)^2 = a^2r^2 = a \cdot ar^2 = ac$.
Q15 — Geometric Progression · easy · theory
The $n$th term of a GP with first term $a$ and common ratio $r$ is:
A. $a + (n-1)r$
B. $ar^{n-1}$ ✓ Correct
C. $ar^n$
D. $a(r^n - 1)$
Solution: The $n$th term of a GP is defined as $T_n = ar^{n-1}$, where $a$ is the first term and $r$ is the common ratio. For $n=1$: $T_1 = ar^0 = a$. For $n=2$: $T_2 = ar$, etc.
Q16 — Geometric Progression · easy · theory
The sum of the first $n$ terms of a GP with first term $a$ and common ratio $r \neq 1$ is:
A. $\frac{a(r^n - 1)}{r - 1}$
B. $\frac{a(1 - r^n)}{1 - r}$
C. Both (A) and (B) ✓ Correct
D. $\frac{ar(r^n - 1)}{r - 1}$
Solution: Both formulas are equivalent: $S_n = \frac{a(r^n - 1)}{r - 1} = \frac{a(1 - r^n)}{1 - r}$. These differ only by a factor of $-1$ in numerator and denominator, which cancel out.
Q17 — Geometric Progression · easy · numerical
If the first term of a GP is $2$ and the common ratio is $3$, then the 5th term is:
A. $2 \cdot 3^4 = 162$ ✓ Correct
B. $2 \cdot 3^5 = 486$
C. $2 + 4 \cdot 3 = 14$
D. $2 \cdot 3^6 = 1458$
Solution: Using $T_n = ar^{n-1}$ with $a = 2$, $r = 3$, and $n = 5$: $T_5 = 2 \cdot 3^{5-1} = 2 \cdot 3^4 = 2 \cdot 81 = 162$.
Q18 — Geometric Progression · easy · theory
The sum of the infinite GP $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots$ is:
A. $1$
B. $2$ ✓ Correct
C. $\frac{1}{2}$
D. $\infty$
Solution: This is an infinite GP with $a = 1$ and $r = \frac{1}{2}$. Since $|r| < 1$, the sum converges to $S = \frac{a}{1-r} = \frac{1}{1 - \frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2$.
Q19 — Geometric Progression · easy · theory
For an infinite GP with first term $a$ and common ratio $r$, the sum exists if and only if:
A. $r > 0$
B. $|r| < 1$ ✓ Correct
C. $r < 1$
D. $r \neq 1$
Solution: The sum $S = \frac{a}{1-r}$ of an infinite GP converges if and only if $|r| < 1$. If $|r| \geq 1$, the series diverges.
Q20 — Geometric Progression · easy · numerical
If $2, x, 18$ are in GP, then $x = $:
A. $6$ or $-6$ ✓ Correct
B. $10$
C. $8$
D. $4$ or $5$
Solution: If $2, x, 18$ are in GP, then $x^2 = 2 \cdot 18 = 36$, so $x = \pm 6$. Both values are valid: for $x = 6$, ratio is $3$; for $x = -6$, ratio is $-3$.
Q21 — Geometric Progression · easy · numerical
The sum of the first 4 terms of a GP with $a = 1$ and $r = 2$ is:
A. $7$
B. $15$ ✓ Correct
C. $31$
D. $8$
Solution: Using $S_n = \frac{a(r^n - 1)}{r - 1}$: $S_4 = \frac{1(2^4 - 1)}{2 - 1} = \frac{16 - 1}{1} = 15$. Verify: $1 + 2 + 4 + 8 = 15$.
Q22 — Geometric Progression · easy · numerical
The sum of the infinite series $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots$ is:
A. $1$ ✓ Correct
B. $\frac{1}{2}$
C. $2$
D. Divergent
Solution: This is a GP with $a = \frac{1}{2}$ and $r = \frac{1}{2}$. The sum is $S = \frac{a}{1-r} = \frac{1/2}{1 - 1/2} = \frac{1/2}{1/2} = 1$.
Q23 — Geometric Progression · easy · numerical
The common ratio of a GP whose first term is $3$ and the sum of infinite terms is $12$ is:
A. $\frac{3}{4}$ ✓ Correct
B. $\frac{1}{4}$
C. $\frac{1}{2}$
D. $\frac{2}{3}$
Solution: Given $a = 3$, $S = 12$, and $|r| < 1$. From $S = \frac{a}{1-r}$: $12 = \frac{3}{1-r} \Rightarrow 12(1-r) = 3 \Rightarrow 12 - 12r = 3 \Rightarrow 12r = 9 \Rightarrow r = \frac{3}{4}$.
Q24 — Geometric Progression · easy · theory
If the 2nd, 3rd, and 4th terms of a GP are $x, y, z$ respectively, then the relationship is:
A. $y^2 = xz$ ✓ Correct
B. $x^2 = yz$
C. $z^2 = xy$
D. $xyz = 1$
Solution: If $a, ar, ar^2, ar^3$ are the first four terms with 2nd, 3rd, 4th being $x, y, z$, then $x = ar, y = ar^2, z = ar^3$. Thus $y^2 = (ar^2)^2 = a^2r^4 = (ar)(ar^3) = xz$.
Q25 — Geometric Progression · easy · numerical
The number of terms in the GP $1, 2, 4, 8, \ldots, 512$ is:
A. $9$
B. $10$ ✓ Correct
C. $11$
D. $12$
Solution: GP: $1, 2, 4, \ldots, 512$ with $a = 1, r = 2$. Last term $512 = 2^9 = ar^{n-1}$, so $2^{n-1} = 2^9 \Rightarrow n = 10$.
Q26 — Arithmetic Progression · hard · theory
If the sum of first $m$ terms of an AP is $n$ and the sum of first $n$ terms is $m$, then the sum of first $(m+n)$ terms is:
A. $-(m+n)$ ✓ Correct
B. $m+n$
C. $mn$
D. $m-n$
Solution: Let $S_m = n$ and $S_n = m$. For an AP, $S_k = \frac{k}{2}[2a + (k-1)d]$. From the given conditions and algebraic manipulation, $S_{m+n} = -(m+n)$.
Q27 — Arithmetic Progression · hard · theory
If the $p$-th, $q$-th, and $r$-th terms of an AP are $a$, $b$, and $c$ respectively, then $(q-r)a + (r-p)b + (p-q)c$ equals:
A. 0 ✓ Correct
B. 1
C. $a+b+c$
D. $p+q+r$
Solution: Let the AP have first term $A$ and common difference $D$. Then $a = A + (p-1)D$, $b = A + (q-1)D$, $c = A + (r-1)D$. The expression $(q-r)a + (r-p)b + (p-q)c$ expands and simplifies to 0.
Q28 — Arithmetic Progression · hard · theory
If the ratio of the sum of $n$ terms of two APs is $(7n + 1) : (4n + 27)$, then the ratio of their $n$-th terms is:
A. $(14n - 6) : (8n + 23)$ ✓ Correct
B. $(7n - 6) : (4n + 23)$
C. $(14n + 6) : (8n - 23)$
D. $(7n + 6) : (4n - 23)$
Solution: For two APs with $\frac{S_n}{S_n'} = \frac{7n+1}{4n+27}$, the ratio of $n$-th terms is $\frac{a_n}{b_n} = \frac{S_{2n-1}}{S_{2n-1}'} = \frac{7(2n-1)+1}{4(2n-1)+27} = \frac{14n - 6}{8n + 23}$.
Q29 — Arithmetic Progression · hard · theory
If the sum of $p$ terms of an AP is $q$ and the sum of $q$ terms is $p$, then the sum of $(p+q)$ terms is:
A. $-(p+q)$ ✓ Correct
B. $p+q$
C. $pq$
D. $p-q$
Solution: Using the same approach as for similar problems, we derive that $S_{p+q} = -(p+q)$.
Q30 — Arithmetic Progression · hard · theory
If the $m$-th term of an AP is $\frac{1}{n}$ and the $n$-th term is $\frac{1}{m}$, then the $mn$-th term is:
A. 1 ✓ Correct
B. $\frac{1}{mn}$
C. $mn$
D. $m+n$
Solution: Let the AP have first term $a$ and common difference $d$. From $a_m = \frac{1}{n}$ and $a_n = \frac{1}{m}$, we get $(m-n)d = \frac{m-n}{mn}$, so $d = \frac{1}{mn}$. Then $a = \frac{1}{mn}$. Thus $a_{mn} = \frac{1}{mn} + (mn-1) \cdot \frac{1}{mn} = 1$.