Sequence and Series JEE Main Questions with Solutions 2027

Sequence and Series JEE Main Questions - Last 5 Years' Most Asked Topics

The analysis of sequence and series JEE Main questions will help the students understand the topics that have come repeatedly in the JEE Main paper. The following is the topic-wise breakdown of questions provided in the source material.

Also Check: JEE Main Maths: 20 Most Repeated Topic-Based Questions 

Sequence and Series JEE Main Questions  With Solutions

Previous year questions form one of the best methods of preparing for the chapter on Sequence and Series. They give an insight to students on the application of concepts and formulas in the JEE Main exam.

The following question bank deals with concepts like Arithmetic Progression (AP), Geometric Progression (GP), Infinite Series, Summation, Arithmetic Means, and Special Series. Students must try to solve all questions independently before referring to the solutions.

Question 1: The sum of the first ten terms of an A.P. is 160 and the sum of the first two terms of a G.P. is 8 . If the first term of the A.P. is equal to the common ratio of the G.P. and the first term of the G.P. is equal to common difference of the A.P., then the sum of all possible values of the first term of the G.P. is:

(1) 349

(2) 3413

(3) 329

(4) 3213

Correct Answer: Option (1)

Solution: 

Sum of 10 terms of A.P, S10=160

$\begin{aligned}
& 5[2 a+9 d]=160 \\
& 2 a+9 d=32 \\
& G . P . \Rightarrow d, d a, d a^2 \ldots \\
& \Rightarrow d+d a=8 \\
& \Rightarrow \frac{d+d(32-9 d)}{2}=8 \\
& \Rightarrow 2 d+32 d-9 d^2=8 \\
& \Rightarrow 9 d^2-34 d+8=0 \\
& d_1+d_2=\frac{34}{9}
\end{aligned}$

Question 2: The value of 13-23+33-.+153 is:

(1) 1706

(2) 1856

(3) 1982

(4) 2403

Correct Answer: Option (2)

Solution: 

$\begin{aligned} & \left(1^3-2^3\right)+\left(3^3-4^3\right)+\ldots .13^3-14^3+15^3 \\ & \text { use } n^3-(n+1)^3=-3 n^2-3 n-1 \\ & \text { put } n=1,3,5, \ldots .13(7 \text { terms }) \\ & \text { sum of pairs }=-1519 \\ & \text { add last term }=15^3 \\ & \Rightarrow 3375-1519=1856\end{aligned}$

Question 3: The sum 1+1212+22+1312+22+32+.. upto 10 terms is equal to:

(1) 130

(2) 155

(3) 3152

(4) 3252

Correct Answer: Option (3)

Solution: $\begin{aligned} & T_n=\frac{1}{n}\left(1^2+2^2+\ldots+n^2\right) \\ & T_n=\frac{n(n+1)(2 n+1)}{6 n} \\ & T_n=\frac{1}{6}\left[2 n^2+3 n+1\right] \\ & S_n=\Sigma T_n=\frac{1}{6}\left[2 \Sigma n^2+3 \Sigma n+\Sigma 1\right] \\ & \Sigma T_n=\frac{1}{6}\left[2 \cdot \frac{n(n+1)(2 n+1)}{6}+\frac{3 n(n+1)}{2}+n\right] \\ & S_{10}=\frac{1}{6}\left[\frac{2 \times 10 \times 11 \times 21}{6}+\frac{3}{2} \times 10 \times 11+10\right] \\ & =\frac{10}{6}\left[77+\frac{33}{2}+1\right]=\frac{315}{2}\end{aligned}$

Question 4: Let =3+4+8+9+13+14+ upto 40 terms. If ()1020 is a root of the equation x2+x-2=0,0,2, then 2+32 is equal to:

(1) 2

(2) 74

(3) 52

(4) 32

Correct Answer: Option (1)

Solution: $\begin{aligned} & a=(3+8+13 \ldots \text { upto } 20 \text { terms })+(4+9+14+ \\ & \ldots \text { upto } 20 \text { terms }) \\ & =\frac{20}{2}[6+19 \times 5]+\frac{20}{2}(8+19 \times 5) \\ & =2040 \\ & (\tan \beta)^{\frac{2040}{1020}}=\tan ^2 \beta \\ & x^2+x-2=0 \int_{-2}^1 \\ & \Rightarrow \tan ^2 \beta=1 ; \sin ^2 \beta=\frac{1}{2} \& \cos ^2 \beta=\frac{1}{2} \\ & \sin ^2 \beta+3 \cos ^2 \beta=\frac{1}{2}+\frac{3}{2}=\frac{4}{2}=2\end{aligned}$

Question 5: Let A1,A2,A3,..,A39 be 39 arithmetic means between the numbers 59 and 159 . Then the mean of A25,~A28,~A31 and A36 is equal to:

(1) 129

(2) 136

(3) 131.50

(4) 134

Correct Answer: Option (4)

Solution: 

$\begin{aligned}
& 59 A_1 A_2 A_3 \ldots A_{39} 159 \\
& d=\frac{159-59}{39+1}=\frac{5}{2}
\end{aligned}$

Now,

$\begin{aligned}
& A_{25}=59+25 d=121.5 \\
& A_{28}=59+28 d=129 \\
& A_{31}=59+31 d=136.5 \\
& A_{36}=59+36 d=149
\end{aligned}$

Mean of

$\begin{aligned}
& \mathrm{A}_{25}, \mathrm{~A}_{28}, \mathrm{~A}_{31}, \mathrm{~A}_{36}=\frac{121.5+129+136.5+149}{4} \\
& =\frac{536}{4}=134
\end{aligned}$

Question 6: If the sum of the first 10 terms of the series 11+144+21+244+31+344+41+444+ is $\frac{\mathrm{m}}{\mathrm{n}}, \operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$, then m+n is equal to:

(1) 256

(2) 264

(3) 276

(4) 284

Correct Answer: Option (3)

Solution: $\begin{aligned} & \mathrm{T}_{\mathrm{r}}=\frac{\mathrm{r}}{1+4 \mathrm{r}^4}=\frac{\mathrm{r}}{4 \mathrm{r}^4+4 \mathrm{r}^2+1-4 \mathrm{r}^2}=\frac{\mathrm{r}}{\left(2 \mathrm{r}^2+1\right)^2-(2 \mathrm{r})^2} \\ & =\frac{\mathrm{r}}{\left(2 \mathrm{r}^2+2 \mathrm{r}+1\right)\left(2 \mathrm{r}^2-2 \mathrm{r}+1\right)} \\ & \mathrm{S}_{10}=\frac{1}{4}\left(\frac{1}{1}-\frac{1}{221}\right)=\frac{55}{221}=\frac{\mathrm{m}}{\mathrm{n}} \\ & \therefore \mathrm{m}+\mathrm{n}=276\end{aligned}$

Question 7: Let the sum of the first n terms of an A.P. be 3n2+5n. Then the sum of squares of the first 10 terms of the A.P. is :

(1) 10220

(2) 12860

(3) 15220

(4) 19780

Correct Answer: Option (3)

Solution: $\begin{aligned} & S_n=3 n^2+5 n \\ & T_n=S_n-S_{n-1} \\ & \left.\Rightarrow T_n=\left(3 n^2+5 n\right)-\left(3(n-1)^2\right)+5(n-1)\right) \\ & \Rightarrow T_n=6 n+2 \\ & \sum_{n=1}^{10} T_n^2=\sum_{n=1}^{10}(6 n+2)^2 \\ & =36 \sum_{n=1}^{10} n^2+24 \sum_{n=1}^{10} n+\sum_{n=1}^{10} 4 \\ & =15220\end{aligned}$

Question 8: From a month of 31 days, 3 different dates are selected at random. If the probability that these dates are in an increasing A.P. is equal to ab, where a,bN and $\operatorname{gcd}(\mathrm{a}, \mathrm{b})=1$, then a+b is equal to ____.

(1) 944

(2) 234

(3) 789

(4) 890

Correct Answer: Option (1)

Solution: 

Total numbers of ways of selecting

$3 \text { numbers }={ }^{31} \mathrm{C}_3=\frac{31 \times 30 \times 29}{3 \times 2 \times 1}=4495$

a,b,c are in A.P.
⇒ either a & c both odd numbers or both even numbers

$\begin{aligned}
& ={ }^{16} \mathrm{C}_2+{ }^{15} \mathrm{C}_2 \\
& =120+105 \\
& =225
\end{aligned}$

Probability =2254495=45899

=a~ba+b=944

Question 9: The sum 131+13+231+3+13+23+331+3+5+ up to 8 terms, is :

(1) 70

(2) 71

(3) 72

(4) 73

Correct Answer: Option (2)

Solution: $\begin{aligned} & \mathrm{T}_{\mathrm{r}}=\frac{1^3+2^3+3^3+\ldots \ldots \mathrm{r}^3}{1+3+5+\ldots . .+(2 \mathrm{r}-1)}=\frac{\left(\frac{\mathrm{r}(\mathrm{r}+1)}{2}\right)^2}{\mathrm{r}^2} \\ & =\frac{\mathrm{r}^2+2 \mathrm{r}+1}{4} \\ & \mathrm{~S}_{\mathrm{n}}=\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{T}_{\mathrm{r}} \\ & \mathrm{S}_{\mathrm{n}}=\frac{1}{4} \sum\left(\mathrm{r}^2+2 \mathrm{r}+1\right) \\ & =\frac{1}{4}\left[\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}+2 \frac{\mathrm{n}(\mathrm{n}+1)}{2}+\mathrm{n}\right] \\ & \mathrm{S}_8=\frac{1}{4}\left[\frac{8 \times 9 \times 17}{6}+8 \times 9+8\right] \\ & =\frac{1}{4}[204+72+8]=71\end{aligned}$

Question 10: If k=1nak=6n3, then k=16ak+1-ak362 is equal to_______.

(1) 91

(2) 23

(3) 89

(4) 78

Correct Answer: Option (1)

Solution: $\begin{aligned} & a_1+a_2+\ldots \ldots+a_n=6 n^3 . \\ & a_1+a_2+\ldots \ldots+a_n+a_{n+1}=6(n+1)^3 \\ & a_{n+1}=6(n+1)^3-6 n^3 \\ & =6((n+1)-n)\left((n+1)^2+n^2+n(n+1)\right) \\ & a_{n+1}=6(1)\left(3 n^2+3 n+1\right) \\ & a_n=6\left(3(n-1)^2+3(n-1)+1\right) \\ & =6\left(3\left(n^2-2 n+1+n-1\right)+1\right) \\ & =6\left(3\left(n^2-n\right)+1\right) \\ & a_n=6\left(3 n^2-3 n+1\right) \\ & \sum_{k=1}^6\left(\frac{6\left(3 k^2+3 k+1\right)-6\left(3 k^2-3 k+1\right)}{36}\right)^2 \\ & \sum_{k=1}^6\left(\frac{36 k}{36}\right)^2=\sum_{k=1}^6 k^2=\frac{6 \times 7 \times 13}{6}=91\end{aligned}$

Question 11: 6326+10.1325+10.2324+10.22323++10.2243 is equal to:

(1) 226

(2) 225

(3) 325

(4) 326

Correct Answer: Option (1)

Solution: $\begin{array}{ll}S=\frac{6}{3^{26}}+\frac{10}{3^{25}}\left[\frac{(6)^{25}-1}{6-1}\right] & S=\frac{6}{3^{26}}+\frac{10}{3^{25}}\left[\frac{6^{25}-1}{5}\right] \\ S=\frac{2}{3^{25}}+2\left[2^{25}-\frac{1}{3^{25}}\right] & S=2^{26}\end{array}$

Question 12: In a G.P., if the product of the first three terms is 27 and the set of all possible values for the sum of its first three terms is R-(a,b), then a2+b2 is equal to:

(1) 90

(2) 89

(3) 78

(4) 80

Correct Answer: Option (1)

Solution: 

Let ar,a,ar first 3 terms of G.P □ Product of them =27

$\begin{aligned}
& a=3 \\
& \mathrm{~S}=\text { sum of first } 3 \text { terms }=\frac{a}{r}+a+a r=a\left(\frac{1}{r}+1+r\right)
\end{aligned}$

$\begin{aligned} & S=3\left(\frac{1}{r}+r+1\right) \\ & \text { If } r<0 \\ & \frac{1}{r}+r \leq-2 \\ & \text { if } r>0 \\ & \frac{1}{r}+r \geq 2\end{aligned}$

$\begin{aligned} & \therefore r+\frac{1}{r} \in(-\infty,-2] \cup[2, \infty) \\ & r+\frac{1}{r}+1 \in(-\infty,-1] \cup[3, \infty) \\ & 3\left(r+\frac{1}{r}+1\right) \in(-\infty,-3] \cup[9, \infty) \\ & \therefore S \in(-\infty,-3] \cup[9, \infty) \\ & S \in R-(-3,9) \\ & a=-3, b=9 \\ & a^2+b^2=(-3)^2+9^2=90\end{aligned}$.

Question 13: If (A-B)A+2C2A=1,A,B,C0,2, then

(1) $\tan A, \tan B, \tan $ are in G.P.

(2) A,B,C are in A.P.

(3) A,C,B are in A.P.

(4) A,C,B are in G.P

Correct Answer: Option (4)

Solution: 

A-BA(1+AB)+1+2A1+2C=1

Let A=x,B=y,C=z

$\begin{aligned}
& \frac{x-y}{x(1+x y)}+\frac{1+\frac{1}{x^2}}{1+\frac{1}{z^2}}=1 \\
& \frac{x-y}{x(1+x y)}+\frac{z^2\left(x^2+1\right)}{x^2 \cdot\left(z^2+1\right)}=1 \\
& x(x-y)\left(z^2+1\right)+z^2\left(x^2+1\right)(1+x y)=x^2(1+x y)\left(z^2+1\right)
\end{aligned}$

After Solving 2C=AB

Question 14: The common difference of the A.P.: a1,a2,am is 13 more than the common difference of the A.P.: b1,b2,bn. If b31=-277,b43=-385, and a78=327, then a1 is equal to:

(1) 19

(2) 21

(3) 16

(4) 24

Correct Answer: Option (1)

Solution: $\begin{aligned} & b_{43}-b_{31}=12 d_b=-108 \\ & \therefore d_b=-9 \\ & d_a=-9+13=4 \\ & a_1=a_{78}-77 d_b=327-77 \times 4=327-308=19\end{aligned}$

Question 15: 13+47+132+1347+4272+133+13247+134272+4373+ upto infinite terms, is equal to:

(1) 52

(2) 65

(3) 43

(4) 74

Correct Answer: Option (1)

Solution: 

$\begin{aligned} & \left(\frac{1}{3}+\frac{4}{3}\right)+\left(\frac{1}{3^2}+\frac{1}{3} \times \frac{4}{7}+\frac{4^2}{7^2}\right)+\left(\frac{1}{3^2}+\frac{1}{3^2} \times \frac{4}{7}+\frac{1}{3} \times \frac{4^2}{7^2}+\frac{4^3}{7^3}\right) \\ & a=\frac{1}{3}, b=\frac{4}{7} \\ & S=(a+b)+\left(a^2+a b+b^2\right)+\left(a^3+a^2 b+a b^2+b^3\right) \\ & =\frac{1}{(a-b)}\left[a^2-b^2+a^3-b^3+a^4-b^4+---\infty\right] \\ & =\frac{a^2+a^3+--\infty}{a-b}-\frac{b^2+b^3+-\infty}{a+b}=\frac{a^2}{(a-b)(1-a)}-\frac{b^2}{(a-b)(1-b)} \\ & =\frac{\frac{1}{9}}{\left(\frac{1}{3}-\frac{4}{7}\right)\left(1-\frac{1}{3}\right)}-\frac{1}{\left(\frac{1}{3}-\frac{4}{7}\right)\left(1-\frac{4}{7}\right)}\end{aligned}$

Question 16: Let a1,a2,a3,a4 be an A.P. of four terms such that each term of the A.P. and its common difference are integers. If a1+a2+a3+a4=48 and a1a2a3a4+l4=361, then the largest term of the A.P. is equal to

(1) 24

(2) 27

(3) 23

(4) 21

Correct Answer: Option (2)

Solution: 

Given a1,a2,a3,a4 are in AP as a-3d,a-d,a+d,a+3d
Given a1+a2+a3+a4=48

$\begin{aligned}
& 4 a=48 \\
& a=12
\end{aligned}$

Given a1a2a3a4+l4=361

$\begin{aligned}
& \left(a^1-9 d^2\right)\left(a^2-d^2\right)+l^4=361 \\
& \left(144-9 d^2\right)\left(144-d^2\right)+l^4=361 \\
& 25 d^4-1440 d^2+144^2=361 \\
& \left(5 d^2-144\right)=19^2 \\
& 5 d^2-144=19 \&-19 \\
& d^2=\frac{163}{5} \quad \& \quad \frac{125}{5}=25 \\
& d=\frac{\sqrt{163}}{5} \& \frac{125}{5}=25 \\
& d=\sqrt{\frac{163}{5}} \quad \& 5
\end{aligned}$

Largest term 12+15=27

Question 17: Consider an A.P : a1,a2,a3,an;a1>0 If a2-a1=-34,an=14a1 and i=1nai=5252 then i=117ai is equal to:

(1) 476

(2) 136

(3) 952

(4) 238

Correct Answer: Option (4)

Solution: 

$\begin{aligned} & a_2-a_1=-\frac{3}{4} \Rightarrow a_1+d-a_1=-\frac{3}{4} \Rightarrow d=-\frac{3}{4} \\ & a_n=\frac{1}{4} a_1 \Rightarrow a_1+(n-1) \cdot\left(-\frac{3}{4}\right)=\frac{1}{4} a_1 \\ & \Rightarrow 4 a_1 3 n+3=a_1 \Rightarrow 3 a_1-3 n+3=0 \Rightarrow a_1-n+1=0 \\ & \sum_{i=l}^n a_i=\frac{525}{2} \Rightarrow \frac{n}{2}\left[2 a_1+(n-1) \cdot\left(-\frac{3}{4}\right)\right]=\frac{525}{2} \Rightarrow \frac{n}{2}\left[a_1+a_n\right]=\frac{525}{2} \\ & 8 n a_1-3 n^2+3 n=525 \Rightarrow \frac{n}{2}\left[a_1+\frac{a_1}{4}\right]=\frac{525}{2} \\ & \Rightarrow a_n n=420 \\ & \Rightarrow a_1\left[a_1+1\right]=420=20 \times 21 \\ & \Rightarrow a_1=20\end{aligned}$

Using the sum formula for the first 17 terms S17 :
$
\begin{gathered}
S_{17}=\frac{17}{2}\left[2 a_1+(17-1) d\right] \\
S_{17}=\frac{17}{2}\left[2(20)+16\left(-\frac{3}{4}\right)\right] \\
S_{17}=\frac{17}{2}[40-12] \\
S_{17}=\frac{17}{2} \times 28=17 \times 14=238
\end{gathered}
$

Question 18: If the sum of the first term of an A.P. is 6 and the sum of its first six terms is 4 , then the sum of its first twelve terms is:

(1) -26

(2) -24

(3) -20

(4) -22

Correct Answer: Option (4)

Solution: 

a+(a+d)+(a+2d)+(d+3d)=6

$\begin{aligned}
& 2 a+3 d=3 \\
& 6 a+15 d=4 \\
& d=-5 / 6 \\
& a=11 / 4 \\
& S_{12}=\frac{12}{2}\left[2\left(\frac{11}{4}\right)+11 \times(-5 / 6)\right]=-22
\end{aligned}$

Hence, the correct answer is option (4).

Question 19: The positive integer n , for which the solutions of the equation x(x+2)+(x+2)(x+4)++(x+2n-2)(x+2n)=8n3 are two consecutive even integers, is:

(1) 3

(2) 9

(3) 6

(4) 12

Correct Answer: Option (1)

Solution: 

$\begin{aligned}
& n x^2+(2 x+6 x+10 x+\ldots \ldots+(4 n-2))+(2.4+4.6+\ldots+(2 n-2) 2 n)=\frac{8 n}{3} \\
& n x^2+2 x(1+3+5+\ldots .+2 n-1)+\sum_{r=1}^n(2 r-2) 2 r=\frac{8 n}{3} \\
& n x^2+2 n^2 x+4\left(\frac{n(n+1)(2 n-2)}{6}\right)=\frac{8 n}{3} \\
& 3 x^2+6 n x+4\left(n^2-1\right)=8
\end{aligned}$

For n=3

$\begin{aligned}
& 3 x^2+18 x+24=0 \\
& x=-2,-4 \text { (two consecutive even integers) }
\end{aligned}$

Question 20: If the sum of the first 10 terms of the series 4.11+4.14+4.21+4.24+4.31+4.34+ is mn, where $\operatorname{gcd}(m, n)=1$, then m+n is equal to______.

(1) 441

(2) 245

(3) 700

(4) 789

Correct Answer: Option (1)

Solution: 

Tr=4r1+4r4

=4r(2r2+2r+1)(2r2-2r+1)

=(2r2+2r+1)-(2r2-2r+1)(2r2+2r+1)(2r2-2r+1)

=12r2-2r+1-12r2+2r+1

Now, sum of first 10 terms S10=r=110 Tr=r=11012r2-2r+1-12r2+2r+1

=11-15+15-113++1181-1221

=1-1221=220221

m=220,n=221

m+n=441

Hence, the answer is 441.

Question 21: Let an be the $\mathrm{n}^{\text {th }}$ term of an A. P.
If Sn=a1+a2+a3++an=700,a6=7 and S7=7, then an is equal to :

(1) 56

(2) 65

(3) 64

(4) 70

Correct Answer: Option (3)

Solution: 

$\begin{aligned}
& a_6=7=a+5 d=7 \\
& S_7=\frac{7}{2}(2 a+6 d)=7(a+3 d)=7 \\
& \Rightarrow a+3 d=1
\end{aligned}
$

Using (i) & (ii)

$\begin{aligned}
& \Rightarrow \quad 2 d=6 \Rightarrow d=3, a_1=-8 \\
& S_n=\frac{n}{2}(2 a+(n-1) d)=700 \\
& =\frac{n}{2}(-16+(n-1) 3)=700 \\
& =n(3 n-19)=1400, n>0 \Rightarrow n=25 \\
& \Rightarrow \quad a_{25}=(-8)+24 \cdot(3)=72-8=64
\end{aligned}$
Hence, the correct answer is option (3).

Question 22: The sum of the infinite series -174+-1194+-1394+-1674+.. is :-

(1) 2+-112

(2) 2--112

(3) 2+-112

(4) 2--112

Correct Answer: Option (4)

Solution: 

Given series:

$S = \cot^{-1}\left(\frac{7}{4}\right) + \cot^{-1}\left(\frac{19}{4}\right) + \cot^{-1}\left(\frac{39}{4}\right) + \cot^{-1}\left(\frac{67}{4}\right) + \cdots
$

Notice the pattern in the arguments:

$\frac{7}{4}, \frac{19}{4}, \frac{39}{4}, \frac{67}{4}, \ldots
$

Rewrite these as:
$
\frac{7}{4} = \frac{4 \cdot 1^2 + 3}{4}, \quad \frac{19}{4} = \frac{4 \cdot 2^2 + 3}{4}, \quad \frac{39}{4} = \frac{4 \cdot 3^2 + 3}{4}, \quad \frac{67}{4} = \frac{4 \cdot 4^2 + 3}{4}, \ldots
$

So the general term is:
$
T_r = \cot^{-1} \left( \frac{4r^2 + 3}{4} \right)
$

Using the identity:
$
\cot^{-1} x = \tan^{-1} \frac{1}{x}
$
we get:
$
T_r = \tan^{-1} \left( \frac{4}{4r^2 + 3} \right)
$

Observe that:
$
\frac{4}{4r^2 + 3} = \frac{1}{r^2 + \frac{3}{4}}
$

Now, write it as:
$
T_r = \tan^{-1} \left( \frac{1}{\left(r + \frac{1}{2}\right)^2 - \left(r - \frac{1}{2}\right)^2} \right)
$

Using the difference of inverse tangent formula:
$
\tan^{-1} a - \tan^{-1} b = \tan^{-1} \left( \frac{a - b}{1 + ab} \right)
$

we identify:
$
T_r = \tan^{-1} \left(r + \frac{1}{2}\right) - \tan^{-1} \left(r - \frac{1}{2}\right)
$

Therefore, the sum of first n terms is telescoping:
$
S_n = \sum_{r=1}^n T_r = \sum_{r=1}^n \left[ \tan^{-1} \left(r + \frac{1}{2}\right) - \tan^{-1} \left(r - \frac{1}{2}\right) \right]
$
$
= \left[\tan^{-1} \left(\frac{3}{2}\right) - \tan^{-1} \left(\frac{1}{2}\right)\right] + \left[\tan^{-1} \left(\frac{5}{2}\right) - \tan^{-1} \left(\frac{3}{2}\right)\right] + \cdots + \left[\tan^{-1} \left(n + \frac{1}{2}\right) - \tan^{-1} \left(n - \frac{1}{2}\right)\right]
$

Most terms cancel, leaving:
$
S_n = \tan^{-1} \left(n + \frac{1}{2}\right) - \tan^{-1} \left(\frac{1}{2}\right)
$

Taking the limit as n:
$
S = \lim_{n \to \infty} S_n = \lim_{n \to \infty} \tan^{-1} \left(n + \frac{1}{2}\right) - \tan^{-1} \left(\frac{1}{2}\right)
$

Since x-1x=2, we have:
$
S = \frac{\pi}{2} - \tan^{-1} \left(\frac{1}{2}\right)
$

Question 23: 1+3+52+7+92+ up to 40 terms is equal to

(1) 43890

(2) 41880

(3) 33980

(4) 40870

Correct Answer: Option (2)

Solution: 

$\left(1^2+5^2+9^2+\ldots . . \text { upto } 20 \text { terms }\right)+(3+7+11+
$... up to 20 terms)

$\begin{aligned}
& =\sum_{\mathrm{r}=1}^{20}(4 \mathrm{r}-3)^2+\sum_{\mathrm{r}=1}^{20}(4 \mathrm{r}-1) \\
& =\sum_{\mathrm{r}=1}^{20}(4 \mathrm{r}-3)^2+(4 \mathrm{r}-1) \\
& =4 \sum_{\mathrm{r}=1}^{20}\left(4 \mathrm{r}^2-5 \mathrm{r}+2\right)
\end{aligned}$
=16r=120r2-20r=120r+8r=1201=41880

Question 24: The sum $1+3+11+25+45+71+........ $ up to 20 terms is equal to

(1) 7240

(2) 7130

(3) 6982

(4) 8124

Correct Answer: Option (1)

Solution: 

Given sum is
$
S_n = 1 + 3 + 11 + 25 + 45 + 71 + \cdots + T_n
$

First order differences are in A.P.

Thus, we can assume that
$
T_n = an^2 + bn + c
$

Solving
$
\begin{cases}
T_1 = 1 = a + b + c \\
T_2 = 3 = 4a + 2b + c \\
T_3 = 11 = 9a + 3b + c
\end{cases}
$

we get a=3,b=-7,c=5

Hence, general term of the given series is:
$
T_n = 3n^2 - 7n + 5
$

Hence, required sum equals:
$
\sum_{n=1}^{20} (3n^2 - 7n + 5)
= 3 \sum_{n=1}^{20} n^2 - 7 \sum_{n=1}^{20} n + 5 \cdot 20
$

Using standard formulas:
$
\sum_{n=1}^{k} n = \frac{k(k+1)}{2}, \quad
\sum_{n=1}^{k} n^2 = \frac{k(k+1)(2k+1)}{6}
$

Apply for k=20:
$
= 3 \cdot \frac{20 \cdot 21 \cdot 41}{6} - 7 \cdot \frac{20 \cdot 21}{2} + 5 \cdot 20
= 7240
$
Hence, the answer is 7240.

Question 25: Consider an A.P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms lies between 1600 and 1800. Then its $11^{\text {th }}$ term is:

(1) 84

(2) 122

(3) 90

(4) 108

Correct Answer: Option (3)

Solution: 

We are given:

- An A.P. of positive integers.
- Sum of first 3 terms $ S_3 = 54 $.
- Sum of first 20 terms $ S_{20} $ lies between 1600 and 1800.

Sum of first 3 terms

S3=32[2a+(3-1)d]=3(a+d)=54

a+d=18

Sum of first 20 terms

S20=202[2a+(20-1)d]=10(2a+19d)

Substitute $ a = 18 - d $:

S20=10[2(18-d)+19d]=10(36-2d+19d)=10(36+17d)

Given:

1600<S20<1800

1600<10(36+17d)<1800

160<36+17d<180

124<17d<144

7517<d<8817

Find integer $ d $

Since $ d $ is a positive integer:

d=8

From $ a + d = 18 $:

a=10

Find the $ 11^{\text{th}} $ term

a11=a+10d=10+108=90

Also Check: 

• JEE Main Physics: 20 Most Repeated Topic-Based Questions

• JEE Main Chemistry: 20 Most Repeated Topic-Based Questions 

JEE Main 2026 Sequence and Series Most Asked Concepts

The table below shows the comparison between the number of questions on various concepts of Sequence and Series in the January and April sessions of JEE Main 2026.

Also Check: JEE Main 2027 3D Geometry: Top Questions to Boost Your Maths Score | JEE Main 2027 Vectors: One-Shot Notes, Formulas & PYQs 

Sequence and Series Important Formulas for JEE Mains

1. nth Term of an AP: `an=a+(n-1)d`

2. Sum of n Terms of an AP: `Sn=n2[2a+(n-1)d]`

3. Sum of an AP Using the Last Term: `Sn=n2(a+l)`

4. nth Term of a GP: `an=arn-1`

5. Sum of n Terms of a GP: `Sn=a(rn-1)r-1, r1`

6. Sum of an Infinite GP: `S=a1-r, |r|<1`

7. Arithmetic Mean of Two Numbers: `A=a+b2`

8. Geometric Mean of Two Positive Numbers: `G=ab

9. Sum of the First n Natural Numbers: `k=1nk=n(n+1)2

10. Sum of Squares of First n Natural Numbers: `k=1nk2=n(n+1)(2n+1)6

11. Sum of Cubes of First n Natural Numbers: `k=1nk3=n(n+1)22

12. Sum of First n Odd Numbers: `1+3+5++(2n-1)=n2`

13. Sum of First n Even Numbers: `2+4+6++2n=n(n+1)

14. Sum of an Arithmetico-Geometric Series: `k=1nkrk-1=1-(n+1)rn+nrn+1(1-r)2, r1

Best Books for JEE  Main Sequence and Series Questions

While preparing Sequence and Series questions for JEE, it is very important to follow the best books because they help properly cover all the concepts. Refer to the Best Books for JEE Main 2027 given below for solving Sequence and Series questions for the JEE PDF. 

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