Sequences and Series JEE Main PYQs (2020–2026)

Solved Sequences and Series PYQs

3 recent questions with answers. The full set of 232 is free to practise.

1. JEE Main 2026EasyMethod of Differences & Telescoping Series (V-N Method)
   Q1

   The sum $1+\frac{1}{2}\left(1^2+2^2\right)+\frac{1}{3}\left(1^2+2^2+3^2\right)+\dots$ upto 10 terms is equal to :

   • A130
   • B155
   • C$\frac{315}{2}$
   • D$\frac{325}{2}$

   Show answer & solution

   Correct answer: (C)

   $$\begin{array}{rl}& T_n=\frac{1}{n}\left(1^2+2^2+\dots +n^2\right)\\ & T_n=\frac{n(n+1)(2n+1)}{6n}\\ & T_n=\frac{1}{6}\left[2n^2+3n+1\right]\\ & S_n=\Sigma T_n=\frac{1}{6}\left[2\Sigma n^2+3\Sigma n+\Sigma 1\right]\\ & \Sigma T_n=\frac{1}{6}\left[2\cdot \frac{n(n+1)(2n+1)}{6}+\frac{3n(n+1)}{2}+n\right]\\ & S_{10}=\frac{1}{6}\left[\frac{2\times 10\times 11\times 21}{6}+\frac{3}{2}\times 10\times 11+10\right]\\ & =\frac{10}{6}\left[77+\frac{33}{2}+1\right]=\frac{315}{2}\end{array}$$
2. JEE Main 2026EasySum of Special Series using Sigma Notation
   Q2

   The value of $1^3-2^3+3^3-\dots +15^3$ is:

   • A1706
   • B1856
   • C1982
   • D2403

   Show answer & solution

   Correct answer: (B)

   We need to find
   
   $$1^3-2^3+3^3-4^3+\cdots +15^3$$
   
   This is an alternating sum of cubes.
   
   Let us group the terms in pairs:
   
   $$\left(1^3-2^3\right)+\left(3^3-4^3\right)+\left(5^3-6^3\right)+$$
   
   $$\cdots +(13^3-14^3)+15^3$$
   
   Now calculate each pair:
   
   $$1^3-2^3=1-8=-7$$
   
   $$3^3-4^3=27-64=-37$$
   
   $$5^3-6^3=125-216=-91$$
   
   $$7^3-8^3=343-512=-169$$
   
   $$9^3-10^3=729-1000=-271$$
   
   $$11^3-12^3=1331-1728=-397$$
   
   $$13^3-14^3=2197-2744=-547$$
   
   Also,
   
   $$15^3=3375$$
   
   Now add all these:
   
   $$-7-37-91-169-271-397-547+3375$$
   
   First add the negative terms:
   
   $$7+37+91+169+271+397+547=1519$$
   
   So,
   
   $$3375-1519=1856$$
   
   Hence, the value is
   
   $$\boxed{1856}$$
   
   So, the correct option is B.
3. JEE Main 2026EasyMeans: A.M., G.M., H.M. and their Inequalities
   Q3

   Let $A_1,A_2,A_3,\dots \dots ..,A_{39}$ be 39 arithmetic means between the numbers 59 and 159. Then the mean of $A_{25},A_{28},A_{31}$ and $A_{36}$ is equal to :

   • A129
   • B136
   • C131.50
   • D134

   Show answer & solution

   Correct answer: (D)

   Since there are $39$ arithmetic means between $59$ and $159$ , the full arithmetic progression has
   
   $$59,A_1,A_2,A_3,\dots ,A_{39},159$$
   
   So total number of terms is
   
   $$39+2=41$$
   
   Hence, the number of intervals is
   
   $$41-1=40$$
   
   Therefore, the common difference is
   
   $$d=\frac{159-59}{40}=\frac{100}{40}=2.5$$
   
   Now the terms are:
   
   $$A_1=59+2.5=61.5$$
   
   In general,
   
   $$A_n=59+n(2.5)$$
   
   Now find the required terms:
   
   $$A_{25}=59+25(2.5)=59+62.5=121.5$$
   
   $$A_{28}=59+28(2.5)=59+70=129$$
   
   $$A_{31}=59+31(2.5)=59+77.5=136.5$$
   
   $$A_{36}=59+36(2.5)=59+90=149$$
   
   Now their mean is
   
   $$\frac{A_{25}+A_{28}+A_{31}+A_{36}}{4}$$
   
   $$=\frac{121.5+129+136.5+149}{4}$$
   
   $$=\frac{536}{4}=134$$
   
   So the correct answer is:
   
   $$\boxed{134}$$
   
   Hence, Option D is correct.

Topic-wise question count

Where JEE Main has asked from Sequences and Series, most frequent first.

• Arithmetic Progression (A.P.) & Properties78
• Geometric Progression (G.P.) & Infinite G.P.52
• Sum of Special Series using Sigma Notation32
• Method of Differences & Telescoping Series (V-N Method)26
• Arithmetic-Geometric Progression (A.G.P.)20
• Means: A.M., G.M., H.M. and their Inequalities13
• Harmonic Progression (H.P.) & Miscellaneous Sequences10

Year-wise Sequences and Series questions in JEE Main

Number of questions from this chapter in each year's paper(s).