Arithmetic Progression (A.P.) — JEE Main Mathematics MCQs with Solutions
Free JEE Main Mathematics Arithmetic Progression (A.P.) MCQs with step-by-step solutions (20 questions). Part of Sequences and Series. Practise online on Prepizo — no login needed.
▶ Practise Arithmetic Progression (A.P.) online (free)
Questions with solutions
Q1 — 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: $T_{10} = a + 9d = 3 + 9(4) = 39$.
Q2 — Arithmetic Progression (A.P.) · medium · numerical
The sum of the first $20$ terms of the A.P. $2, 5, 8, \dots$ is:
A. $620$
B. $600$
C. $610$ ✓ Correct
D. $590$
Solution: $S_{20} = \tfrac{20}{2}[2(2) + 19(3)] = 10(4 + 57) = 610$.
Q3 — Arithmetic Progression (A.P.) · medium · numerical
Which term of the A.P. $4, 9, 14, \dots$ is $104$?
A. $20$-th
B. $21$-st ✓ Correct
C. $19$-th
D. $22$-nd
Solution: $4 + 5(n-1) = 104 \Rightarrow 5(n-1) = 100 \Rightarrow n = 21$.
Q4 — Arithmetic Progression (A.P.) · medium · numerical
In an A.P. the $7$-th term is $34$ and the $13$-th term is $64$. The common difference is:
A. $6$
B. $3$
C. $5$ ✓ Correct
D. $4$
Solution: $(a+12d)-(a+6d) = 64-34 \Rightarrow 6d = 30 \Rightarrow d = 5$.
Q5 — Arithmetic Progression (A.P.) · medium · numerical
The number of terms in the A.P. $7, 13, 19, \dots, 205$ is:
A. $34$ ✓ Correct
B. $32$
C. $33$
D. $35$
Solution: $7 + 6(n-1) = 205 \Rightarrow 6(n-1) = 198 \Rightarrow n = 34$.
Q6 — 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 $S_{15} = 15 \times 20 = 300$.
Q7 — Arithmetic Progression (A.P.) · medium · numerical
If $S_n = 2n^2 + 3n$ for an A.P., then its $10$-th term is:
A. $39$
B. $43$
C. $40$
D. $41$ ✓ Correct
Solution: $T_n = S_n - S_{n-1} = 4n + 1$, so $T_{10} = 41$.
Q8 — Arithmetic Progression (A.P.) · medium · numerical
Three numbers in A.P. have sum $15$ and product $80$. The numbers are:
A. $4, 5, 6$
B. $1, 5, 9$
C. $2, 5, 8$ ✓ Correct
D. $3, 5, 7$
Solution: Take $5-d, 5, 5+d$ (sum $15 \Rightarrow$ middle $5$); $5(25-d^2)=80 \Rightarrow d^2 = 9 \Rightarrow d = 3$.
Q9 — Arithmetic Progression (A.P.) · medium · numerical
If four arithmetic means are inserted between $3$ and $23$, the third mean is:
A. $19$
B. $11$
C. $13$
D. $15$ ✓ Correct
Solution: With $6$ terms, $d = \tfrac{23-3}{5} = 4$; the means are $7, 11, 15, 19$, so the third is $15$.
Q10 — 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 $T_{p+q}=a+(p+q-1)d=0$.
Q11 — Arithmetic Progression (A.P.) · medium · theory
If each term of an A.P. with common difference $d$ is multiplied by a constant $k$, the new sequence is:
A. an A.P. with common difference $kd$ ✓ Correct
B. a G.P. with ratio $k$
C. an A.P. with the same common difference
D. not a progression
Solution: Multiplying every term by $k$ preserves the A.P. structure and scales the common difference to $kd$.
Q12 — Arithmetic Progression (A.P.) · medium · numerical
The sum of the first $n$ odd natural numbers $1 + 3 + 5 + \cdots + (2n-1)$ for $n = 20$ is:
A. $400$ ✓ Correct
B. $380$
C. $420$
D. $441$
Solution: The sum of the first $n$ odd numbers is $n^2 = 20^2 = 400$.
Q13 — 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 $6m + 2 = 164 \Rightarrow m = 27$.
Q14 — Arithmetic Progression (A.P.) · medium · numerical
If the sum of $n$ terms of an A.P. is $S_n = n^2$, its common difference is:
A. $4$
B. $1$
C. $2$ ✓ Correct
D. $3$
Solution: $T_n = S_n - S_{n-1} = 2n - 1$; the terms $1, 3, 5, \dots$ have $d = 2$.
Q15 — 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 $40, 60, 80$.
Q16 — 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 $\tfrac{a+20d}{a+4d} = \tfrac{24d}{8d} = 3:1$.
Q17 — Arithmetic Progression (A.P.) · medium · numerical
The arithmetic mean of $12$ and $30$ is:
A. $24$
B. $20$
C. $21$ ✓ Correct
D. $18$
Solution: A.M. $= \tfrac{12+30}{2} = 21$.
Q18 — 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 $m=11$, $2m-1 = 21$: $\tfrac{148}{111}$.
Q19 — Arithmetic Progression (A.P.) · medium · numerical
If $a, b, c$ are in A.P., then $2b$ equals:
A. $a - c$
B. $a^2 + c^2$
C. $a + c$ ✓ Correct
D. $ac$
Solution: For an A.P., the middle term is the average: $b = \tfrac{a+c}{2}$, i.e. $2b = a + c$.
Q20 — 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 $= \tfrac{30}{2}(12+99) = 15 \times 111 = 1665$.