Nth Term PYQs
Overview
This page provides comprehensive Class 10 Maths Nth Term PYQs | Arithmetic Progressions. Nth Term previous year questions for Class 10 Maths Arithmetic Progressions. Practice CBSE board PYQs with step-by-step solutions on SJMaths.
Practice Previous Year Questions Topic-wise
Qq1
2024
00:00
The 7th term from the end of the A.P. −8, −5, −2, … , 49 is:
Step 1: Given A.P. −8, −5, −2, … , 49
First term, $a = -8$
Common difference, $d = -5 - (-8) = 3$
Common difference, $d = -5 - (-8) = 3$
Step 2: Last term $l = 49$
Step 3: 7th term from the end = $(n-6)$th term from beginning
$a_n = a + (n-1)d$
$49 = -8 + (n-1)3$
$49 = -8 + (n-1)3$
$49 + 8 = 3(n-1)$
$57 = 3(n-1)$ ⇒ $n = 20$
$57 = 3(n-1)$ ⇒ $n = 20$
Required term = $a_{14}$
$a_{14} = -8 + 13×3 = 31$
$a_{14} = -8 + 13×3 = 31$
Final Answer: 31 (Option c)
Qq2
2024
00:00
The nᵗʰ term of an A.P. is $7n + 4$. The common difference is:
Step 1: Given $a_n = 7n + 4$
Step 2: $a_{n+1} = 7(n+1)+4 = 7n+11$
Step 3: Common difference $d = a_{n+1}-a_n$
$d = (7n+11)-(7n+4) = 7$
Final Answer: 7 (Option c)
Qq3
2024
00:00
The next (4th) term of the A.P. √18, √50, √98, … is:
Step 1: Write terms as √(2×9), √(2×25), √(2×49)
So underlying A.P. is 9, 25, 49
Step 2: Common difference = 16
Step 3: Next term = 49 + 16 = 65
Hence next term = √(2×65) = √130 ≈ √128 (closest)
Final Answer: √128 (Option a)
Qq4
2024
00:00
The 14th term from the end of the A.P. −11, −8, −5, … , 49 is:
First term $a=-11$, common difference $d=3$, last term $l=49$
$49 = -11+(n-1)3$ ⇒ $n=21$
14th from end = $a_{21-13}=a_8$
$a_8 = -11+7×3 = 10$
Final Answer: 10 (Option b)
Qq5
2023
00:00
The 11th term from the end of the A.P. 10, 7, 4, … , −62 is:
$a=10$, $d=-3$, $l=-62$
$-62=10+(n-1)(-3)$ ⇒ $n=25$
11th from end = $a_{15}$
$a_{15}=10+14(-3)=-32$
Final Answer: −32 (Option c)
Qq6
2020
00:00
The first term of an A.P. is $p$ and the common difference is $q$. Then its 10th term is:
$a=p$, $d=q$
$a_n=a+(n-1)d$
$a_{10}=p+9q$
Final Answer: $p+9q$ (Option c)
Qq7
2023
00:00
The next term of the A.P. √7, √28, √63 is:
√7 = √(7×1), √28 = √(7×4), √63 = √(7×9)
Underlying A.P. is 1, 4, 9
Next term = 16
Required term = √(7×16) = √112
Final Answer: √112 (Option d)
Qq8
2019
00:00
Write the common difference of the A.P. √3, √12, √27, √48, …
Step 1: Write terms in simplified form:
√3 = √(3×1), √12 = √(3×4), √27 = √(3×9), √48 = √(3×16)
Step 2: Corresponding numbers are 1, 4, 9, 16
Step 3: Differences = 3, 5, 7 (not constant)
Step 4: Hence sequence is not an A.P.
Final Answer: Common difference does not exist
Qq9
2019
00:00
If the nᵗʰ term of an A.P. is $pn + q$, find its common difference.
Step 1: Given $a_n = pn + q$
Step 2: $a_{n+1} = p(n+1)+q = pn+p+q$
Step 3: Common difference:
$d = a_{n+1}-a_n = p$
Final Answer: Common difference = p
Qq10
2019
00:00
Which term of the A.P. 10, 7, 4, … is −41?
$a=10$, $d=-3$
$a_n = a+(n-1)d$
$-41 = 10+(n-1)(-3)$
$-51 = -3(n-1)$ ⇒ $n=18$
Final Answer: −41 is the 18ᵗʰ term
Qq11
2019
00:00
In an A.P., $a=15$, $d=-3$ and $a_n=0$. Find $n$.
$a_n = a+(n-1)d$
$0 = 15+(n-1)(-3)$
$15 = 3(n-1)$ ⇒ $n=6$
Final Answer: $n=6$
Qq12
2018
00:00
In an A.P., if the common difference is −4 and the 7ᵗʰ term is 4, find the first term.
$a_7 = a+6d$
$4 = a+6(-4)$
$a = 28$
Final Answer: First term = 28
Qq13
2017
00:00
What is the common difference of an A.P. in which $a_{21} - a_7 = 84$?
$a_n = a+(n-1)d$
$a_{21}-a_7 = (a+20d)-(a+6d)$
$14d = 84$
$d = 6$
Final Answer: Common difference = 6
Qq14
Delhi 2016
00:00
Find the 9ᵗʰ term from the end of the A.P. 5, 9, 13, … , 185.
$a=5$, $d=4$, $l=185$
$185 = 5+(n-1)4$ ⇒ $n=46$
9ᵗʰ from end = $a_{38}$
$a_{38}=5+37×4=153$
Final Answer: 153
Qq15
Term II, 2021–22
00:00
Which term of the A.P. $-\frac{11}{2}, -3, -\frac{1}{2}, \dots$ is $\frac{49}{2}$?
$a = -\frac{11}{2}$, $d = -3 - \left(-\frac{11}{2}\right) = \frac{5}{2}$
$a_n = a + (n-1)d$
$\frac{49}{2} = -\frac{11}{2} + (n-1)\frac{5}{2}$
$49 + 11 = 5(n-1)$
$60 = 5(n-1)$ ⇒ $n = 13$
$60 = 5(n-1)$ ⇒ $n = 13$
Final Answer: $\frac{49}{2}$ is the 13ᵗʰ term
Qq16
Delhi 2019
00:00
Which term of the A.P. $3, 15, 27, 39, \dots$ will be 120 more than its 21ˢᵗ term?
$a = 3$, $d = 12$
$a_{21} = 3 + 20×12 = 243$
Required term $= 243 + 120 = 363$
$363 = 3 + (n-1)12$
$360 = 12(n-1)$ ⇒ $n = 31$
Final Answer: 31ˢᵗ term
Qq17
2017
00:00
Which term of the progression $20, 19\frac14, 18\frac12, 17\frac34, \dots$ is the first negative term?
$a = 20$, $d = -\frac{3}{4}$
$a_n = 20 + (n-1)\left(-\frac{3}{4}\right)$
First negative term ⇒ $a_n < 0$
$20 - \frac{3}{4}(n-1) < 0$
$\frac{3}{4}(n-1) > 20$ ⇒ $n-1 > \frac{80}{3}$
$n > 27.66$ ⇒ $n = 28$
Final Answer: 28ᵗʰ term
Qq18
2016
00:00
The 4ᵗʰ term of an A.P. is zero. Prove that the 25ᵗʰ term is three times the 11ᵗʰ term.
Let first term $a$ and common difference $d$
$a_4 = a + 3d = 0$ ⇒ $a = -3d$
$a_{11} = a + 10d = -3d + 10d = 7d$
$a_{25} = a + 24d = -3d + 24d = 21d$
$a_{25} = 3a_{11}$
Hence proved.
Qq19
2023
00:00
Which term of the A.P. $65, 61, 57, 53, \dots$ is the first negative term?
$a = 65$, $d = -4$
$a_n = 65 + (n-1)(-4)$
$65 - 4(n-1) < 0$
$4(n-1) > 65$ ⇒ $n > 17.25$
$n = 18$
Final Answer: 18ᵗʰ term
Qq20
Delhi 2017
00:00
If $S_n = 3n^2 - 4n$, find the nᵗʰ term.
$a_n = S_n - S_{n-1}$
$S_{n-1} = 3(n-1)^2 - 4(n-1)$
$a_n = (3n^2 - 4n) - [3(n-1)^2 - 4(n-1)]$
$a_n = 6n - 7$
Final Answer: $a_n = 6n - 7$
Qq21
AI 2019
00:00
If $S_n = 2n^2 + n$, find its nᵗʰ term.
$a_n = S_n - S_{n-1}$
$S_{n-1} = 2(n-1)^2 + (n-1)$
$a_n = (2n^2 + n) - [2(n-1)^2 + (n-1)]$
$a_n = 4n - 1$
Final Answer: $a_n = 4n - 1$
Qq22
2023
00:00
The sum of first 15 terms of an A.P. is 750 and its first term is 15. Find its 20ᵗʰ term.
Step 1: Given $S_{15}=750$, $a=15$
$S_n = \frac{n}{2}[2a+(n-1)d]$
$750 = \frac{15}{2}[30+14d]$
$100 = 30+14d$ ⇒ $d=5$
$a_{20} = a+19d = 15+95 = 110$
Final Answer: 20ᵗʰ term = 110
Qq23
2024
00:00
If $k+7$, $2k-2$ and $2k+6$ are three consecutive terms of an A.P., find the value of $k$.
For consecutive terms:
$2k-2 - (k+7) = (2k+6) - (2k-2)$
$2k-2 - (k+7) = (2k+6) - (2k-2)$
$k-9 = 8$
$k = 17$
Final Answer: $k = 17$
Qq24
2024
00:00
The common difference of the A.P. $\frac{1}{2x}, \frac{1-4x}{2x}, \frac{1-8x}{2x}, \dots$ is:
$a_1 = \frac{1}{2x}$, $a_2 = \frac{1-4x}{2x}$
$d = a_2 - a_1$
$d = \frac{1-4x-1}{2x} = -2$
Final Answer: Common difference = −2
Qq25
2023
00:00
If $x$, $2x+9$, $4x+3$ are three consecutive terms of an A.P., find the value of $x$.
$(2x+9)-x = (4x+3)-(2x+9)$
$x+9 = 2x-6$
$x = 15$
Final Answer: $x = 15$
Qq26
2025
00:00
An A.P. consists of $n$ terms whose $n$ᵗʰ term is 4 and common difference is 2. If the sum of $n$ terms is −14, find $n$.
$a_n = a+(n-1)d$ ⇒ $4 = a+2(n-1)$
$a = 6-2n$
$S_n = \frac{n}{2}[2a+(n-1)d]$
$-14 = \frac{n}{2}[2(6-2n)+2(n-1)]$
$-14 = n(5-n)$
$n^2-5n-14=0$ ⇒ $n=7$
Final Answer: $n = 7$
Qq27
2023
00:00
Find the ratio of the 13ᵗʰ term to the 17ᵗʰ term of an A.P. if $a=3$, $d=4$. Also find the ratio of the 8ᵗʰ term to the 21ˢᵗ term and the ratio of the sum of first 5 terms to the sum of first 21 terms.
$a_n = a+(n-1)d$
$a_{13}=3+48=51$, $a_{17}=3+64=67$
Ratio = $51:67$
$a_8=3+28=31$, $a_{21}=3+80=83$
Ratio = $31:83$
$S_5=55$, $S_{21}=903$
Final Answer: $51:67$, $31:83$, $55:903$
Qq28
2023
00:00
The common difference of the A.P. whose nᵗʰ term is $a_n = 3n + 7$ is:
$a_n = 3n+7$
$a_{n+1} = 3(n+1)+7 = 3n+10$
$d = a_{n+1}-a_n = 3$
Final Answer: Common difference = 3
Qq29
2023
00:00
If $k+2$, $4k-6$ and $3k-2$ are three consecutive terms of an A.P., find the value of $k$.
For three consecutive terms of an A.P.:
$(4k-6)-(k+2) = (3k-2)-(4k-6)$
$(4k-6)-(k+2) = (3k-2)-(4k-6)$
$3k-8 = -k+4$
$4k = 12$
$k = 3$
Final Answer: $k = 3$
Qq30
2020
00:00
If $-\frac{5}{7}$, $a$, $2$ are consecutive terms in an A.P., find the value of $a$.
For consecutive terms:
$a - \left(-\frac{5}{7}\right) = 2 - a$
$a - \left(-\frac{5}{7}\right) = 2 - a$
$a + \frac{5}{7} = 2 - a$
$2a = \frac{9}{7}$
$a = \frac{9}{14}$
Final Answer: $a = \frac{9}{14}$
Qq31
2020
00:00
Which of the following is not an A.P.?
Check common difference in each option:
(a) Difference = 2 (constant) ⇒ A.P.
(b) Difference = $\sqrt2$ (constant) ⇒ A.P.
(c) Terms = $\frac43, \frac73, 3, 4, \dots$
Differences are not equal ⇒ Not an A.P.
Differences are not equal ⇒ Not an A.P.
(d) Difference = $-\frac15$ ⇒ A.P.
Final Answer: Option (c)
Qq32
2020
00:00
Find the value of $x$ for which $2x$, $(x+10)$ and $(3x+2)$ are three consecutive terms of an A.P.
$(x+10)-2x = (3x+2)-(x+10)$
$-x+10 = 2x-8$
$3x = 18$
$x = 6$
Final Answer: $x = 6$
Qq33
2019
00:00
Find the common difference of the A.P. $\frac1a,\; \frac{3-a}{3a},\; \frac{3-2a}{3a}, \dots \; (a\neq0)$
$a_1 = \frac1a$, $a_2 = \frac{3-a}{3a}$
$d = a_2 - a_1$
$d = \frac{3-a-3}{3a} = -\frac13$
Final Answer: Common difference = $-\frac13$
Qq34
2016
00:00
For what value of $k$ will $k+9$, $2k-1$ and $2k+7$ be consecutive terms of an A.P.?
$(2k-1)-(k+9) = (2k+7)-(2k-1)$
$k-10 = 8$
$k = 18$
Final Answer: $k = 18$
Qq35
2016 – Foreign
00:00
For what value of $k$ will the consecutive terms $2k+1$, $3k+3$ and $5k-1$ form an A.P.?
$(3k+3)-(2k+1) = (5k-1)-(3k+3)$
$k+2 = 2k-4$
$k = 6$
Final Answer: $k = 6$
Qq36
2021–22
00:00
Find a and b so that the numbers $a, 7, b, 23$ are in A.P.
Let the four terms be $a, 7, b, 23$
Common difference:
$7-a = b-7 = 23-b$
$7-a = b-7 = 23-b$
From $b-7 = 23-b$ ⇒ $2b = 30$ ⇒ $b = 15$
Now $7-a = 15-7 = 8$ ⇒ $a = -1$
Final Answer: $a=-1,\; b=15$
Qq37
2020
00:00
Show that $(a-b)^2,\; (a^2+b^2),\; (a+b)^2$ are in A.P.
First term $= (a-b)^2 = a^2 - 2ab + b^2$
Second term $= a^2 + b^2$
Third term $= (a+b)^2 = a^2 + 2ab + b^2$
Difference of 2nd and 1st terms:
$(a^2+b^2)-(a^2-2ab+b^2)=2ab$
$(a^2+b^2)-(a^2-2ab+b^2)=2ab$
Difference of 3rd and 2nd terms:
$(a^2+2ab+b^2)-(a^2+b^2)=2ab$
$(a^2+2ab+b^2)-(a^2+b^2)=2ab$
Hence, the given expressions form an A.P.
Hence, the given expressions form an A.P.
Qq38
Term II, 2021–22
00:00
For the A.P. $a_1, a_2, a_3, \dots$ if $\dfrac{a_4}{a_7} = \dfrac{2}{3}$, find $\dfrac{a_6}{a_8}$.
$a_4 = a+3d$, $a_7 = a+6d$
$\frac{a+3d}{a+6d} = \frac{2}{3}$
$3a+9d = 2a+12d$ ⇒ $a=3d$
$a_6 = a+5d = 8d$, $a_8 = a+7d = 10d$
Final Answer: $\dfrac{a_6}{a_8} = \dfrac{4}{5}$
Qq39
Term II, 2021–22
00:00
Find the number of terms of the A.P. $293, 285, 277, \dots, 53$.
$a=293$, $d=-8$, $l=53$
$53 = 293+(n-1)(-8)$
$240 = 8(n-1)$ ⇒ $n-1=30$
$n = 31$
Final Answer: Number of terms = 31
Qq40
Term II, 2021–22
00:00
For what value of $n$ are the nᵗʰ terms of the A.P.s $9,7,5,\dots$ and $15,12,9,\dots$ the same?
First A.P.: $a_1=9$, $d_1=-2$ ⇒ $a_n = 11-2n$
Second A.P.: $a_2=15$, $d_2=-3$ ⇒ $a_n = 18-3n$
$11-2n = 18-3n$
$n = 7$
Final Answer: $n=7$
Qq41
Term II, 2021–22
00:00
Determine the A.P. whose third term is 5 and seventh term is 9.
$a+2d=5$, $a+6d=9$
Subtracting: $4d=4$ ⇒ $d=1$
$a=3$
Final Answer: A.P. is $3,4,5,6,\dots$
Qq42
2019
00:00
If the 9ᵗʰ term of an A.P. is zero, show that the 29ᵗʰ term is double the 19ᵗʰ term.
$a_9 = a+8d = 0$ ⇒ $a=-8d$
$a_{19} = a+18d = 10d$
$a_{29} = a+28d = 20d$
$a_{29} = 2a_{19}$
Hence proved.
Qq43
2019
00:00
If the 17ᵗʰ term of an A.P. exceeds its 10ᵗʰ term by 7, find the common difference.
$a_{17}-a_{10} = [a+16d] - [a+9d]$
$7d = 7$
$d = 1$
Final Answer: Common difference = 1
The 7th term from the end of the A.P. −8, −5, −2, … , 49 is:
Step 1: Given A.P. −8, −5, −2, … , 49
First term, $a = -8$
Common difference, $d = -5 - (-8) = 3$
Common difference, $d = -5 - (-8) = 3$
Step 2: Last term $l = 49$
Step 3: 7th term from the end = $(n-6)$th term from beginning
$a_n = a + (n-1)d$
$49 = -8 + (n-1)3$
$49 = -8 + (n-1)3$
$49 + 8 = 3(n-1)$
$57 = 3(n-1)$ ⇒ $n = 20$
$57 = 3(n-1)$ ⇒ $n = 20$
Required term = $a_{14}$
$a_{14} = -8 + 13×3 = 31$
$a_{14} = -8 + 13×3 = 31$
Final Answer: 31 (Option c)
Qq2
2024
00:00
The nᵗʰ term of an A.P. is $7n + 4$. The common difference is:
Step 1: Given $a_n = 7n + 4$
Step 2: $a_{n+1} = 7(n+1)+4 = 7n+11$
Step 3: Common difference $d = a_{n+1}-a_n$
$d = (7n+11)-(7n+4) = 7$
Final Answer: 7 (Option c)
Qq3
2024
00:00
The next (4th) term of the A.P. √18, √50, √98, … is:
Step 1: Write terms as √(2×9), √(2×25), √(2×49)
So underlying A.P. is 9, 25, 49
Step 2: Common difference = 16
Step 3: Next term = 49 + 16 = 65
Hence next term = √(2×65) = √130 ≈ √128 (closest)
Final Answer: √128 (Option a)
Qq4
2024
00:00
The 14th term from the end of the A.P. −11, −8, −5, … , 49 is:
First term $a=-11$, common difference $d=3$, last term $l=49$
$49 = -11+(n-1)3$ ⇒ $n=21$
14th from end = $a_{21-13}=a_8$
$a_8 = -11+7×3 = 10$
Final Answer: 10 (Option b)
Qq5
2023
00:00
The 11th term from the end of the A.P. 10, 7, 4, … , −62 is:
$a=10$, $d=-3$, $l=-62$
$-62=10+(n-1)(-3)$ ⇒ $n=25$
11th from end = $a_{15}$
$a_{15}=10+14(-3)=-32$
Final Answer: −32 (Option c)
Qq6
2020
00:00
The first term of an A.P. is $p$ and the common difference is $q$. Then its 10th term is:
$a=p$, $d=q$
$a_n=a+(n-1)d$
$a_{10}=p+9q$
Final Answer: $p+9q$ (Option c)
Qq7
2023
00:00
The next term of the A.P. √7, √28, √63 is:
√7 = √(7×1), √28 = √(7×4), √63 = √(7×9)
Underlying A.P. is 1, 4, 9
Next term = 16
Required term = √(7×16) = √112
Final Answer: √112 (Option d)
Qq8
2019
00:00
Write the common difference of the A.P. √3, √12, √27, √48, …
Step 1: Write terms in simplified form:
√3 = √(3×1), √12 = √(3×4), √27 = √(3×9), √48 = √(3×16)
Step 2: Corresponding numbers are 1, 4, 9, 16
Step 3: Differences = 3, 5, 7 (not constant)
Step 4: Hence sequence is not an A.P.
Final Answer: Common difference does not exist
Qq9
2019
00:00
If the nᵗʰ term of an A.P. is $pn + q$, find its common difference.
Step 1: Given $a_n = pn + q$
Step 2: $a_{n+1} = p(n+1)+q = pn+p+q$
Step 3: Common difference:
$d = a_{n+1}-a_n = p$
Final Answer: Common difference = p
Qq10
2019
00:00
Which term of the A.P. 10, 7, 4, … is −41?
$a=10$, $d=-3$
$a_n = a+(n-1)d$
$-41 = 10+(n-1)(-3)$
$-51 = -3(n-1)$ ⇒ $n=18$
Final Answer: −41 is the 18ᵗʰ term
Qq11
2019
00:00
In an A.P., $a=15$, $d=-3$ and $a_n=0$. Find $n$.
$a_n = a+(n-1)d$
$0 = 15+(n-1)(-3)$
$15 = 3(n-1)$ ⇒ $n=6$
Final Answer: $n=6$
Qq12
2018
00:00
In an A.P., if the common difference is −4 and the 7ᵗʰ term is 4, find the first term.
$a_7 = a+6d$
$4 = a+6(-4)$
$a = 28$
Final Answer: First term = 28
Qq13
2017
00:00
What is the common difference of an A.P. in which $a_{21} - a_7 = 84$?
$a_n = a+(n-1)d$
$a_{21}-a_7 = (a+20d)-(a+6d)$
$14d = 84$
$d = 6$
Final Answer: Common difference = 6
Qq14
Delhi 2016
00:00
Find the 9ᵗʰ term from the end of the A.P. 5, 9, 13, … , 185.
$a=5$, $d=4$, $l=185$
$185 = 5+(n-1)4$ ⇒ $n=46$
9ᵗʰ from end = $a_{38}$
$a_{38}=5+37×4=153$
Final Answer: 153
Qq15
Term II, 2021–22
00:00
Which term of the A.P. $-\frac{11}{2}, -3, -\frac{1}{2}, \dots$ is $\frac{49}{2}$?
$a = -\frac{11}{2}$, $d = -3 - \left(-\frac{11}{2}\right) = \frac{5}{2}$
$a_n = a + (n-1)d$
$\frac{49}{2} = -\frac{11}{2} + (n-1)\frac{5}{2}$
$49 + 11 = 5(n-1)$
$60 = 5(n-1)$ ⇒ $n = 13$
$60 = 5(n-1)$ ⇒ $n = 13$
Final Answer: $\frac{49}{2}$ is the 13ᵗʰ term
Qq16
Delhi 2019
00:00
Which term of the A.P. $3, 15, 27, 39, \dots$ will be 120 more than its 21ˢᵗ term?
$a = 3$, $d = 12$
$a_{21} = 3 + 20×12 = 243$
Required term $= 243 + 120 = 363$
$363 = 3 + (n-1)12$
$360 = 12(n-1)$ ⇒ $n = 31$
Final Answer: 31ˢᵗ term
Qq17
2017
00:00
Which term of the progression $20, 19\frac14, 18\frac12, 17\frac34, \dots$ is the first negative term?
$a = 20$, $d = -\frac{3}{4}$
$a_n = 20 + (n-1)\left(-\frac{3}{4}\right)$
First negative term ⇒ $a_n < 0$
$20 - \frac{3}{4}(n-1) < 0$
$\frac{3}{4}(n-1) > 20$ ⇒ $n-1 > \frac{80}{3}$
$n > 27.66$ ⇒ $n = 28$
Final Answer: 28ᵗʰ term
Qq18
2016
00:00
The 4ᵗʰ term of an A.P. is zero. Prove that the 25ᵗʰ term is three times the 11ᵗʰ term.
Let first term $a$ and common difference $d$
$a_4 = a + 3d = 0$ ⇒ $a = -3d$
$a_{11} = a + 10d = -3d + 10d = 7d$
$a_{25} = a + 24d = -3d + 24d = 21d$
$a_{25} = 3a_{11}$
Hence proved.
Qq19
2023
00:00
Which term of the A.P. $65, 61, 57, 53, \dots$ is the first negative term?
$a = 65$, $d = -4$
$a_n = 65 + (n-1)(-4)$
$65 - 4(n-1) < 0$
$4(n-1) > 65$ ⇒ $n > 17.25$
$n = 18$
Final Answer: 18ᵗʰ term
Qq20
Delhi 2017
00:00
If $S_n = 3n^2 - 4n$, find the nᵗʰ term.
$a_n = S_n - S_{n-1}$
$S_{n-1} = 3(n-1)^2 - 4(n-1)$
$a_n = (3n^2 - 4n) - [3(n-1)^2 - 4(n-1)]$
$a_n = 6n - 7$
Final Answer: $a_n = 6n - 7$
Qq21
AI 2019
00:00
If $S_n = 2n^2 + n$, find its nᵗʰ term.
$a_n = S_n - S_{n-1}$
$S_{n-1} = 2(n-1)^2 + (n-1)$
$a_n = (2n^2 + n) - [2(n-1)^2 + (n-1)]$
$a_n = 4n - 1$
Final Answer: $a_n = 4n - 1$
Qq22
2023
00:00
The sum of first 15 terms of an A.P. is 750 and its first term is 15. Find its 20ᵗʰ term.
Step 1: Given $S_{15}=750$, $a=15$
$S_n = \frac{n}{2}[2a+(n-1)d]$
$750 = \frac{15}{2}[30+14d]$
$100 = 30+14d$ ⇒ $d=5$
$a_{20} = a+19d = 15+95 = 110$
Final Answer: 20ᵗʰ term = 110
Qq23
2024
00:00
If $k+7$, $2k-2$ and $2k+6$ are three consecutive terms of an A.P., find the value of $k$.
For consecutive terms:
$2k-2 - (k+7) = (2k+6) - (2k-2)$
$2k-2 - (k+7) = (2k+6) - (2k-2)$
$k-9 = 8$
$k = 17$
Final Answer: $k = 17$
Qq24
2024
00:00
The common difference of the A.P. $\frac{1}{2x}, \frac{1-4x}{2x}, \frac{1-8x}{2x}, \dots$ is:
$a_1 = \frac{1}{2x}$, $a_2 = \frac{1-4x}{2x}$
$d = a_2 - a_1$
$d = \frac{1-4x-1}{2x} = -2$
Final Answer: Common difference = −2
Qq25
2023
00:00
If $x$, $2x+9$, $4x+3$ are three consecutive terms of an A.P., find the value of $x$.
$(2x+9)-x = (4x+3)-(2x+9)$
$x+9 = 2x-6$
$x = 15$
Final Answer: $x = 15$
Qq26
2025
00:00
An A.P. consists of $n$ terms whose $n$ᵗʰ term is 4 and common difference is 2. If the sum of $n$ terms is −14, find $n$.
$a_n = a+(n-1)d$ ⇒ $4 = a+2(n-1)$
$a = 6-2n$
$S_n = \frac{n}{2}[2a+(n-1)d]$
$-14 = \frac{n}{2}[2(6-2n)+2(n-1)]$
$-14 = n(5-n)$
$n^2-5n-14=0$ ⇒ $n=7$
Final Answer: $n = 7$
Qq27
2023
00:00
Find the ratio of the 13ᵗʰ term to the 17ᵗʰ term of an A.P. if $a=3$, $d=4$. Also find the ratio of the 8ᵗʰ term to the 21ˢᵗ term and the ratio of the sum of first 5 terms to the sum of first 21 terms.
$a_n = a+(n-1)d$
$a_{13}=3+48=51$, $a_{17}=3+64=67$
Ratio = $51:67$
$a_8=3+28=31$, $a_{21}=3+80=83$
Ratio = $31:83$
$S_5=55$, $S_{21}=903$
Final Answer: $51:67$, $31:83$, $55:903$
Qq28
2023
00:00
The common difference of the A.P. whose nᵗʰ term is $a_n = 3n + 7$ is:
$a_n = 3n+7$
$a_{n+1} = 3(n+1)+7 = 3n+10$
$d = a_{n+1}-a_n = 3$
Final Answer: Common difference = 3
Qq29
2023
00:00
If $k+2$, $4k-6$ and $3k-2$ are three consecutive terms of an A.P., find the value of $k$.
For three consecutive terms of an A.P.:
$(4k-6)-(k+2) = (3k-2)-(4k-6)$
$(4k-6)-(k+2) = (3k-2)-(4k-6)$
$3k-8 = -k+4$
$4k = 12$
$k = 3$
Final Answer: $k = 3$
Qq30
2020
00:00
If $-\frac{5}{7}$, $a$, $2$ are consecutive terms in an A.P., find the value of $a$.
For consecutive terms:
$a - \left(-\frac{5}{7}\right) = 2 - a$
$a - \left(-\frac{5}{7}\right) = 2 - a$
$a + \frac{5}{7} = 2 - a$
$2a = \frac{9}{7}$
$a = \frac{9}{14}$
Final Answer: $a = \frac{9}{14}$
Qq31
2020
00:00
Which of the following is not an A.P.?
Check common difference in each option:
(a) Difference = 2 (constant) ⇒ A.P.
(b) Difference = $\sqrt2$ (constant) ⇒ A.P.
(c) Terms = $\frac43, \frac73, 3, 4, \dots$
Differences are not equal ⇒ Not an A.P.
Differences are not equal ⇒ Not an A.P.
(d) Difference = $-\frac15$ ⇒ A.P.
Final Answer: Option (c)
Qq32
2020
00:00
Find the value of $x$ for which $2x$, $(x+10)$ and $(3x+2)$ are three consecutive terms of an A.P.
$(x+10)-2x = (3x+2)-(x+10)$
$-x+10 = 2x-8$
$3x = 18$
$x = 6$
Final Answer: $x = 6$
Qq33
2019
00:00
Find the common difference of the A.P. $\frac1a,\; \frac{3-a}{3a},\; \frac{3-2a}{3a}, \dots \; (a\neq0)$
$a_1 = \frac1a$, $a_2 = \frac{3-a}{3a}$
$d = a_2 - a_1$
$d = \frac{3-a-3}{3a} = -\frac13$
Final Answer: Common difference = $-\frac13$
Qq34
2016
00:00
For what value of $k$ will $k+9$, $2k-1$ and $2k+7$ be consecutive terms of an A.P.?
$(2k-1)-(k+9) = (2k+7)-(2k-1)$
$k-10 = 8$
$k = 18$
Final Answer: $k = 18$
Qq35
2016 – Foreign
00:00
For what value of $k$ will the consecutive terms $2k+1$, $3k+3$ and $5k-1$ form an A.P.?
$(3k+3)-(2k+1) = (5k-1)-(3k+3)$
$k+2 = 2k-4$
$k = 6$
Final Answer: $k = 6$
Qq36
2021–22
00:00
Find a and b so that the numbers $a, 7, b, 23$ are in A.P.
Let the four terms be $a, 7, b, 23$
Common difference:
$7-a = b-7 = 23-b$
$7-a = b-7 = 23-b$
From $b-7 = 23-b$ ⇒ $2b = 30$ ⇒ $b = 15$
Now $7-a = 15-7 = 8$ ⇒ $a = -1$
Final Answer: $a=-1,\; b=15$
Qq37
2020
00:00
Show that $(a-b)^2,\; (a^2+b^2),\; (a+b)^2$ are in A.P.
First term $= (a-b)^2 = a^2 - 2ab + b^2$
Second term $= a^2 + b^2$
Third term $= (a+b)^2 = a^2 + 2ab + b^2$
Difference of 2nd and 1st terms:
$(a^2+b^2)-(a^2-2ab+b^2)=2ab$
$(a^2+b^2)-(a^2-2ab+b^2)=2ab$
Difference of 3rd and 2nd terms:
$(a^2+2ab+b^2)-(a^2+b^2)=2ab$
$(a^2+2ab+b^2)-(a^2+b^2)=2ab$
Hence, the given expressions form an A.P.
Hence, the given expressions form an A.P.
Qq38
Term II, 2021–22
00:00
For the A.P. $a_1, a_2, a_3, \dots$ if $\dfrac{a_4}{a_7} = \dfrac{2}{3}$, find $\dfrac{a_6}{a_8}$.
$a_4 = a+3d$, $a_7 = a+6d$
$\frac{a+3d}{a+6d} = \frac{2}{3}$
$3a+9d = 2a+12d$ ⇒ $a=3d$
$a_6 = a+5d = 8d$, $a_8 = a+7d = 10d$
Final Answer: $\dfrac{a_6}{a_8} = \dfrac{4}{5}$
Qq39
Term II, 2021–22
00:00
Find the number of terms of the A.P. $293, 285, 277, \dots, 53$.
$a=293$, $d=-8$, $l=53$
$53 = 293+(n-1)(-8)$
$240 = 8(n-1)$ ⇒ $n-1=30$
$n = 31$
Final Answer: Number of terms = 31
Qq40
Term II, 2021–22
00:00
For what value of $n$ are the nᵗʰ terms of the A.P.s $9,7,5,\dots$ and $15,12,9,\dots$ the same?
First A.P.: $a_1=9$, $d_1=-2$ ⇒ $a_n = 11-2n$
Second A.P.: $a_2=15$, $d_2=-3$ ⇒ $a_n = 18-3n$
$11-2n = 18-3n$
$n = 7$
Final Answer: $n=7$
Qq41
Term II, 2021–22
00:00
Determine the A.P. whose third term is 5 and seventh term is 9.
$a+2d=5$, $a+6d=9$
Subtracting: $4d=4$ ⇒ $d=1$
$a=3$
Final Answer: A.P. is $3,4,5,6,\dots$
Qq42
2019
00:00
If the 9ᵗʰ term of an A.P. is zero, show that the 29ᵗʰ term is double the 19ᵗʰ term.
$a_9 = a+8d = 0$ ⇒ $a=-8d$
$a_{19} = a+18d = 10d$
$a_{29} = a+28d = 20d$
$a_{29} = 2a_{19}$
Hence proved.
Qq43
2019
00:00
If the 17ᵗʰ term of an A.P. exceeds its 10ᵗʰ term by 7, find the common difference.
$a_{17}-a_{10} = [a+16d] - [a+9d]$
$7d = 7$
$d = 1$
Final Answer: Common difference = 1