Sequence-series - JEE Advanced Previous Year Questions with Solutions
Sequence and Series JEE Advanced previous year questions cover Arithmetic Progression (AP), Geometric Progression (GP), Harmonic Progression (HP), and AM–GM inequalities. Questions are typically 3–4 marks, single or multiple correct, and test conceptual depth over formula recall. Detailed solutions for all 10 questions are provided below.
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Q. If the sum of first $n$ terms of an A.P. is $\mathrm{c} n^{2},$ then the sum of squares of these $n$ terms is (A) $\frac{\mathrm{n}\left(4 \mathrm{n}^{2}-1\right) \mathrm{c}^{2}}{6}$(B) $\frac{\mathrm{n}\left(4 \mathrm{n}^{2}+1\right) \mathrm{c}^{2}}{3}$(C) $\frac{\mathrm{n}\left(4 \mathrm{n}^{2}-1\right) \mathrm{c}^{2}}{3}$(D) $\frac{n\left(4 n^{2}+1\right) c^{2}}{6}$[JEE 2009, 3 (–1)]
Ans. (C) $\mathrm{C}$$\mathrm{S}_{\mathrm{n}}=\mathrm{cn}^{2}$$\mathrm{S}_{\mathrm{n}-1}=\mathrm{c}(\mathrm{n}-1)^{2}$$\mathrm{T}_{\mathrm{n}}=\mathrm{S}_{\mathrm{n}}-\mathrm{S}_{\mathrm{n}-1}=\mathrm{c}(2 \mathrm{n}-1)$$\mathrm{T}_{\mathrm{n}}^{\prime}=\mathrm{T}_{\mathrm{n}}^{2}=\mathrm{c}^{2}\left(4 \mathrm{n}^{2}-4 \mathrm{n}+1\right)$$\sum \mathrm{T}_{\mathrm{n}}^{\prime}=\mathrm{nc}^{2}\left(\frac{4 \mathrm{n}^{2}-1}{3}\right)$
Q. Let $a_{1}, a_{2}, a_{3}, \ldots \ldots \ldots a_{11}$ be real numbers satisfying $\mathrm{a}_{1}=15,27-2 \mathrm{a}_{2}>0$ and $\mathrm{a}_{\mathrm{k}}=2 \mathrm{a}_{\mathrm{k}-1}-\mathrm{a}_{\mathrm{k}-2}$ for $\mathrm{k}=3,4 \ldots \ldots \ldots 11$If $\frac{a_{1}^{2}+a_{2}^{2}+\ldots \ldots+a_{11}^{2}}{11}=90,$ then the value of $\frac{a_{1}+a_{2}+\ldots \ldots+a_{11}}{11}$ is equal to[JEE 2010,3+3]
Ans. 0
Q. The minimum value of the sum of real numbers $a^{-5}, a^{-4}, 3 a^{-3}, 1, a^{8}$ and $a^{10}$ with $a>0$ is [JEE 2011, 4]
Ans. 8 As $a>0$and all the given terms are positivehence considering A.M. $\geq$ G.M. for given numbers :$\frac{\mathrm{a}^{-5}+\mathrm{a}^{-4}+\mathrm{a}^{-3}+\mathrm{a}^{-3}+\mathrm{a}^{-3}+\mathrm{a}^{8}+\mathrm{a}^{10}}{7} \geq\left(\mathrm{a}^{-5} \cdot \mathrm{a}^{-4} \cdot \mathrm{a}^{-3} \cdot \mathrm{a}^{-3} \cdot \mathrm{a}^{-3} \cdot \mathrm{a}^{10}\right)^{\frac{1}{7}}$$\Rightarrow \frac{\mathrm{a}^{-5}+\mathrm{a}^{-4}+\mathrm{a}^{-3}+\mathrm{a}^{-3}+\mathrm{a}^{-3}+\mathrm{a}^{8}+\mathrm{a}^{10}}{7} \geq 1 \quad \Rightarrow\left(\mathrm{a}^{-5}+\mathrm{a}^{-4}+3 \mathrm{a}^{-3}+\mathrm{a}^{-3}+\mathrm{a}^{10}\right)_{\min }=7$where $\mathrm{a}^{-5}=\mathrm{a}^{-4}=\mathrm{a}^{-3}=\mathrm{a}^{8}=\mathrm{a}^{10}$ i.e. $\mathrm{a}=1$$\Rightarrow\left(\mathrm{a}^{-5}+\mathrm{a}^{-4}+3 \mathrm{a}^{-3}+\mathrm{a}^{8}+\mathrm{a}^{10}+1\right)_{\min }=8 \quad$ when $\mathrm{a}=1$
Q. Let $a_{1}, a_{2}, a_{3}, \ldots \ldots \ldots, a_{100}$ be an arithmetic progression with $a_{1}=3$ and $S_{p}=\sum_{i=1}^{p} a_{i}, 1 \leq p \leq 100 .$ For any integer $n$ with $1 \leq n \leq 20,$ let $m=5 n .$ If $\frac{S_{m}}{S_{n}}$ does not depend on n, then $a_{2}$ is [JEE 2011, 4]
Ans. 9 or 3 9 or 3(Comment : The information about the common difference i.e. zero or non-zero is not given in the question. Hence there are two possible answers)Consider $\mathrm{d} \neq 0$ the solution is$\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots \ldots \ldots \ldots ., \mathrm{a}_{100} \rightarrow \mathrm{AP}$$\mathrm{a}_{1}=3 ; \quad \mathrm{S}_{\mathrm{p}}=\sum_{\mathrm{i}=1}^{\mathrm{p}} \mathrm{a}_{\mathrm{i}} \quad 1 \leq \mathrm{n} \leq 20$$\mathrm{m}=5 \mathrm{n}$$\frac{\mathrm{S}_{\mathrm{m}}}{\mathrm{S}_{\mathrm{n}}}=\frac{\frac{\mathrm{m}}{2}\left[2 \mathrm{a}_{1}+(\mathrm{m}-1) \mathrm{d}\right]}{\frac{\mathrm{n}}{2}\left[2 \mathrm{a}_{1}+(\mathrm{n}-1) \mathrm{d}\right]}$$\frac{\mathrm{S}_{\mathrm{m}}}{\mathrm{S}_{\mathrm{n}}}=\frac{5\left[\left(2 \mathrm{a}_{1}-\mathrm{d}\right)+5 \mathrm{nd}\right]}{\left[\left(2 \mathrm{a}_{1}-\mathrm{d}\right)+\mathrm{nd}\right]}$ for $\frac{\mathrm{S}_{\mathrm{m}}}{\mathrm{S}_{\mathrm{n}}}$ to be independent of $\mathrm{n}$$\begin{array}{ll}{\therefore 2 \mathrm{a}_{1}-\mathrm{d}=0} & {\Rightarrow \mathrm{d}=2 \mathrm{a}_{1} \Rightarrow \mathrm{d}=6 \Rightarrow \mathrm{a}_{2}=9} \\ {\text { If } \mathrm{d}=0 \quad \Rightarrow} & {\mathrm{a}_{2}=\mathrm{a}_{1}=3}\end{array}$
Q. Let $a_{1}, a_{2}, a_{3}, \ldots .$ be in harmonic progression with $a_{1}=5$ and $a_{20}=25 .$ The least positive integer n for which $a_{n}<0$ is (A) 22 (B) 23 (C) 24 (D) 25[JEE 2012, 3 (–1)]
Ans. (D) $a_{1}, a_{2}, a_{3} \dots \ldots . .$ be in $\mathrm{H.P} \Rightarrow \frac{1}{\mathrm{a}_{1}}, \frac{1}{\mathrm{a}_{2}}, \frac{1}{\mathrm{a}_{3}} \ldots$ be in A.P.in A.P. $\mathrm{T}_{1}=\frac{1}{\mathrm{a}_{1}}=\frac{1}{5}$ and $\mathrm{T}_{20}=\frac{1}{\mathrm{a}_{20}}=\frac{1}{25}$$\therefore \mathrm{T}_{20}=\mathrm{T}_{1}+19 \mathrm{d}$$\frac{1}{25}=\frac{1}{5}+19 \mathrm{d} \Rightarrow \mathrm{d}=-\frac{4}{19 \times 25}$$\mathrm{T}_{\mathrm{n}}=\mathrm{T}_{1}+(\mathrm{n}-1) \mathrm{d}<0 \Rightarrow \frac{1}{5}-\frac{(\mathrm{n}-1) \cdot 4}{19 \times 25}<0 \quad \Rightarrow \frac{1}{5}<\frac{4(\mathrm{n}-1)}{25 \times 19} \Rightarrow \frac{5 \times 19}{4}+1<\mathrm{n} \Rightarrow \frac{99}{4}<\mathrm{n}$$\Rightarrow$ least positive integer $n$ is 25
Q. Let $\mathrm{S}_{\mathrm{n}}=\sum_{\mathrm{k}=1}^{4 \mathrm{n}}(-1)^{\frac{\mathrm{k}(\mathrm{k}+1)}{2}} \mathrm{k}^{2} .$ Then $\mathrm{S}_{\mathrm{n}}$ can take value(s) (A) 1056 (B) 1088 (C) 1120 (D) 1332[JEE-Advanced 2013, 4, (–1)]
Ans. (A,D) $\mathrm{S}_{\mathrm{n}}=-1^{2}-2^{2}+3^{2}+4^{2}-5^{2}-6^{2}+7^{2}+8^{2} \ldots \ldots \ldots$$\mathrm{S}_{\mathrm{n}}^{\mathrm{n}}=\left(3^{2}-1^{2}\right)+\left(4^{2}-2^{2}\right)+\ldots \ldots \ldots \ldots$$\mathrm{S}_{\mathrm{n}}=2(1+2+3+\ldots \ldots+4 \mathrm{n})=\frac{2(4 n)(4 \mathrm{n}+1)}{2}$$\mathrm{S}_{\mathrm{n}}=4 \mathrm{n}(4 \mathrm{n}+1)$$\mathrm{S}_{\mathrm{n}}=4 \mathrm{n}(4 \mathrm{n}+1)=1056$ is possible when $\mathrm{n}=8$$4 n(4 n+1)=1088$ not possible$4 n(4 n+1)=1120$ not possible$4 n(4 n+1)=1332$ possible when $n=9$
Q. A pack contains n cards numbered from 1 to n. Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is 1224. If the smaller to the numbers on the removed cards is k, then k – 20 = [JEE-Advanced 2013, 4, (–1)]
Ans. 5 When 1 and 2 are removed from numbers 1 to n then we get maximum possible sum of remaining numbers and when n –1,n are removed then get minimum possible sum of remaining numbers.
Q. Let a, b, c be positive integers such that $\frac{b}{a}$ is an integer. If $a, b, c$ are in geometric progression and the arithmetic mean of $a, b, c$ is $b+2,$ then the value of $\frac{a^{2}+a-14}{a+1}$ is [JEE(Advanced)-2014, 3]
Ans. 4
Q. Let $b_{i}>1$ for $i=1,2, \ldots \ldots, 101 .$ Suppose $\log _{e} b_{1}, \log _{e} b_{2}, \ldots \ldots, \log b_{10}$ are in Arithmetic Progression (A.P.) with the common difference log $2 .$ Suppose $a_{1}, a_{2}, \ldots . ., a_{101}$ are in A.P. such that $a_{1}=$ $b_{1}$ and $a_{51}=b_{51}$. If $t=b_{1}+b_{2}+\ldots . .+b_{51}$ and $s=a_{1}+a_{2}+\ldots . .+a_{51}$ then (A) $\mathrm{s}>\mathrm{t}$ and $\mathrm{a}_{101}>\mathrm{b}_{101}$(B) $\mathrm{s}>\mathrm{t}$ and $\mathrm{a}_{101}<\mathrm{b}_{101}$(C) $\mathrm{s}<\mathrm{t}$ and $\mathrm{a}_{101}>\mathrm{b}_{101}$(D) $\mathrm{s}<\mathrm{t}$ and $\mathrm{a}_{101}<\mathrm{b}_{101}$[JEE(Advanced)-2016]
Ans. B
Q. The sides of the right angled triangle are in arithmetic progression. If the triangle has area 24, then what is the length of its smallest side ? [JEE(Advanced)-2017]
Ans. 6
Q. Let $X$ be the set consisting of the first 2018 terms of the arithmetic progression $1,6,11, \ldots .$ , and $Y$ be the set consisting of the first 2018 terms of the arithmetic progression $9,16,23,$ $\ldots \ldots$ Then, the number of elements in the set $X \cup Y$ is [JEE(Advanced)-2018]
What is the fastest way to identify whether a sequence is AP, GP, or HP?
Check differences first: if $a_{k} - a_{k-1}$ is constant, it is an AP. If $a_k / a_{k-1}$ is constant, it is a GP. If the reciprocals $1/a_k$ form an AP, it is an HP. In JEE Advanced, the recurrence relation form (like $a_k = 2a_{k-1} - a_{k-2}$) always signals an AP — rearrange it as $a_k - a_{k-1} = a_{k-1} - a_{k-2}$.
Is Sequence and Series more important for JEE Main or JEE Advanced?
Both exams test this topic, but differently. JEE Main tests direct formula application — sum formulas, nth term, inserting means. JEE Advanced tests conceptual depth, pattern recognition, and multi-step reasoning, often combining AP/GP/HP with inequalities or number theory. You need a stronger conceptual grip for JEE Advanced
How many questions from Sequence and Series appear in JEE Advanced each year?
Typically, 1–2 questions appear from Sequence and Series in JEE Advanced each year. The topic carries around 4–8 marks per paper. Between 2009 and 2023, there has been at least one question from this chapter in every paper. It is considered a medium-difficulty but high-reward topic because the concepts are well-defined.
Where can I find more Sequence and Series practice material for JEE Advanced?
After completing these PYQs, practise from the NCERT Solutions for Class 11 Maths Chapter 9 for foundational problems, then move to eSaral's topic-wise test series. eSaral's JEE Advanced course, taught by IIT Bombay faculty including AIR-41 rankers, includes chapter-wise video lectures, solved examples, and a 5-layer doubt-solving system to clear every conceptual block.
What is the standard method to solve HP problems in JEE Advanced?
Always convert the HP to its corresponding AP by taking reciprocals. Find the common difference of the AP using the given terms, then work entirely within the AP. Convert back to HP only for the final answer. There is no direct sum formula for HP, so every HP JEE Advanced question is really an AP problem in disguise.
How do I apply AM–GM in sequence problems where coefficients are greater than 1?
Split the term with a coefficient into that many equal separate terms before applying AM–GM. For example, $3a^{-3}$ becomes three separate $a^{-3}$ terms. This ensures the product of all terms simplifies cleanly (usually to 1), allowing you to find the exact minimum. This technique appeared directly in the JEE Advanced 2011 paper.
Comments
Aryan Gupta
Aug. 26, 2024, 6:35 a.m.
Pw zinda baad
Payel
April 12, 2023, 6:35 a.m.
Just Only few questions are available.
Aryan gupta
Aug. 26, 2024, 6:35 a.m.
Heyy Payal
atriavo
May 6, 2021, 5:58 p.m.
good
atriavo
May 6, 2021, 5:57 p.m.
out
Ankur
Nov. 17, 2020, 11:02 p.m.
Very good explanation
Nikhil Malhotra
Oct. 22, 2020, 9:29 p.m.
Good question..
Hots questions
Hardik joshi
Oct. 22, 2020, 12:14 p.m.
E saral waale bhaiyya ya sir mene aapko yt pe dekha hai aur in questions me kuch questions nahi dikhre saaf se..
Wo do lines ek ke upr ek aari hai unme to plzz is ko theek krdo aap
Sarthak Mahajan
Oct. 7, 2020, 3:11 p.m.
Very helpful questions
sam
Sept. 16, 2020, 11:21 p.m.
gracias
TEJESH
July 28, 2020, 6:38 p.m.
All the PYQs are not there...Your content is seriously very limited
