Sequences and Series JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Sequences and Series, free to read — no sign-in needed. The full chapter has 32 questions; sign in to attempt the remaining 27 in the exam simulator.

1. Q1JEE Advanced Adv 2025 (Paper 1)
   Let $\mathbb{R}$ denote the set of all real numbers. Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function such that $f(x)\gt 0$ for all $x \in \mathbb{R}$, and $f(x+y)=f(x) f(y)$ for all $x, y \in \mathbb{R}$. Let the real numbers $\mathrm{a}_1, \mathrm{a}_2 \ldots, \mathrm{a}_{50}$ be in an arithmetic progression. If $f\left(\mathrm{a}_{31}\right)=64 f\left(\mathrm{a}_{25}\right)$, and $\sum_{i=1}^{50} f\left(a_i\right)=3\left(2^{25}+1\right)$ then the value of $\sum_{i=6}^{30} f\left(a_i\right)$ is ________
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   Answer: 96

   $\begin{aligned} & \because f(x+y)=f(x) \cdot f(y) \\ & \Rightarrow f(x)=k^x \quad(f(x)\gt 0 \forall x \in R) \\ & \because f\left(a_{31}\right)=64 f\left(a_{25}\right)\end{aligned}$ $\begin{aligned} & \Rightarrow \mathrm{k}^{(\mathrm{a}+30 \mathrm{~d})}=64 \cdot \mathrm{k}^{(\mathrm{a}+24 \mathrm{~d})} \\ & \Rightarrow \mathrm{k}^{6 \mathrm{~d}}=64 \\ & \Rightarrow \mathrm{k}^{\mathrm{d}}=2\end{aligned}$ $\sum_{\mathrm{i}=1}^{50} \mathrm{f}\left(\mathrm{a}_{\mathrm{i}}\right)=\mathrm{f}\left(\mathrm{a}_1\right)+\mathrm{f}\left(\mathrm{a}_2\right)+\ldots \ldots+\mathrm{f}\left(\mathrm{a}_{50}\right)$ $\begin{aligned} & =\mathrm{k}^{\mathrm{a}}+\mathrm{k}^{\mathrm{a}+\mathrm{d}}+\ldots+\mathrm{k}^{\mathrm{a}+49 \mathrm{~d}}=\frac{\mathrm{k}^{\mathrm{a}}\left(\mathrm{k}^{50 \mathrm{~d}}-1\right)}{\mathrm{k}^{\mathrm{d}}-1} \\ & =\mathrm{k}^{\mathrm{a}}\left(2^{50}-1\right)=3\left(2^{25}+1\right)(\text { Given }) \\ & \Rightarrow \mathrm{k}^{\mathrm{a}}=\frac{3}{2^{25}-1}\end{aligned}$ $\begin{aligned} & \therefore \sum_{\mathrm{i}=6}^{30} \mathrm{f}\left(\mathrm{a}_{\mathrm{i}}\right)=\mathrm{k}^{\mathrm{a}+5 \mathrm{~d}}+\mathrm{k}^{\mathrm{a}+6 \mathrm{~d}}+\ldots+\mathrm{k}^{\mathrm{a}+29 \mathrm{~d}} \\ & \quad=\mathrm{k}^{\mathrm{a}+5 \mathrm{~d}} \frac{\left(\mathrm{k}^{25 \mathrm{~d}}-1\right)}{\mathrm{k}^{\mathrm{d}}-1}=\mathrm{k}^{\mathrm{a}} \cdot\left(\mathrm{k}^{\mathrm{d}}\right)^5\left(2^{25}-1\right) \\ & \quad=\frac{3}{2^{25}-1} \cdot 2^5\left(2^{25}-1\right)=96\end{aligned}$
2. Q2JEE Advanced Adv 2023 (Paper 1)
   Let $7\overbrace{5\ldots 5}7^{r}$ denote the $\left(r+2\right)$ digit number where the first and the last digits are $7$ and the remaining $r$ digits are $5$. Consider the sum $S=77+757+7557+\ldots +7\overbrace{5\ldots 5}7^{98}$. If $S=\dfrac{7\overbrace{5\ldots 5}^{99}7+m}{n}$, where $m$ and $n$ are natural numbers less than $3000$, then the value of $m+n$ is
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   Answer: 1219

   Given, $S=77+757+7557+\ldots +7\overbrace{5\ldots 5}7^{98}$ $\Rightarrow S=7\times 10+7+7\times 100+5\times 10+7+7\times 1000+5\times 100+5\times 10+7\ldots +7\overbrace{5\ldots 5}7^{98}$ $\Rightarrow S=7\left(10+{10}^{2}+\ldots +{10}^{99}\right)+50\left(1+11+\ldots +\overbrace{111\ldots 1}^{98}\right)+7\times 99$ $\Rightarrow S=70\left(\dfrac{{10}^{99}-1}{9}\right)+\dfrac{50}{9}\left[\left(10-1\right)+\left({10}^{2}-1\right)+\ldots +\left({10}^{98}-1\right)\right]+7\times 99$ $\Rightarrow S=70\left(\dfrac{{10}^{99}-1}{9}\right)+\dfrac{50}{9}\left[10\left(\dfrac{{10}^{98}-1}{9}\right)-98\right]+7\times 99$ $\Rightarrow S=\dfrac{7\times {10}^{100}}{9}-\dfrac{70}{9}+\dfrac{50}{9}\left[\dfrac{{10}^{99}-1-9}{9}-98\right]+7\times 99$ $\Rightarrow S=\dfrac{7\times {10}^{100}}{9}-\dfrac{70}{9}+\dfrac{50}{9}\left[\overbrace{111\ldots 1}^{99}-99\right]+7\times 99$ $\Rightarrow S=\dfrac{7\times {10}^{100}-70+\overbrace{555\ldots 5}^{99}0}{9}-550+693$ $\Rightarrow S=\dfrac{7\overbrace{555\ldots .5}^{99}-70+143\times 9}{9}$ $\Rightarrow S=\dfrac{7\overbrace{55\ldots 5}^{99}7+1210}{9}$ So, on comparing we get, $m+n=1210+9=1219$
3. Q3JEE Advanced Adv 2022 (Paper 1)
   Let ${a}_{1},{a}_{2},{a}_{3},\ldots$ be an arithmetic progression with ${a}_{1}=7$ and common difference $8$. Let ${T}_{1},{T}_{2},{T}_{3},\ldots$ be such that ${T}_{1}=3$ and ${T}_{n+1}-{T}_{n}={a}_{n}$ fo $n\geq 1$. Then, which of the following is/are TRUE?
   1. A.${T}_{20}=1604$
   2. B.$\sum _{k=1}^{20}{T}_{k}=10510$
   3. C.${T}_{30}=3454$
   4. D.$\sum _{k=1}^{30}{T}_{k}=35610$
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   Answer: B,C

   Given, ${a}_{n}=7+\left(n-1\right)8$ and ${T}_{1}=3$ Also ${T}_{n+1}={T}_{n}+{a}_{n}$ Now ${T}_{n}={T}_{n-1}+{a}_{n-1}$ Putting $n=1$ we get, ${T}_{2}={T}_{1}+{a}_{1}$ Now putting $n=2$ we get, ${T}_{3}={T}_{2}+{a}_{2}\Rightarrow {T}_{3}={T}_{1}+{a}_{1}+{a}_{2}$ And so on we get, $\Rightarrow {T}_{n+1}={T}_{1}+{a}_{1}+{a}_{2}+\ldots +{a}_{n}$ $\Rightarrow {T}_{n+1}={T}_{1}+\dfrac{n}{2}\left[2\left(7\right)+\left(n-1\right)8\right]$ $\Rightarrow {T}_{n+1}={T}_{1}+n\left(4n+3\right)\cdots \left(1\right)$ Now for $n=19$ we get, ${T}_{20}=3+\left(19\right)\left(79\right)=1504$ And for $n=29$ we get, ${T}_{30}=3+\left(29\right)\left(119\right)=3454$ {option C is correct} Now finding sum we get, $\sum _{k=1}^{20}{T}_{k}=3+\sum _{k=2}^{20}{T}_{k}=3+\sum _{k=1}^{19}\left(3+4{n}^{2}+3n\right)$ $=3+3\left(19\right)+\dfrac{3\left(19\right)\left(20\right)}{2}+\dfrac{4\left(19\right)\left(20\right)\left(39\right)}{6}$ $=3+10507=10510$ {option B is correct} And Similarly $\sum _{k=1}^{30}{T}_{k}=3+\sum _{k=1}^{29}\left(4{n}^{2}+3n+3\right)=35615$
4. Q4JEE Advanced Adv 2022 (Paper 1)
   Let ${l}_{1},{l}_{2}...{l}_{100}$ be consecutive terms of an arithmetic progression with common difference ${d}_{1}$, and let ${w}_{1},{w}_{2}$, $\ldots ,{w}_{100}$ be consecutive terms of another arithmetic progression with common difference ${d}_{2}$, where ${d}_{1}{d}_{2}=10$. For each $i=1,2,3..........100$, let ${R}_{i}$ be a rectangle with length ${l}_{i}$, width ${w}_{i}$ and area ${A}_{i}$.If ${A}_{51}-{A}_{50}=1000$, then the value of ${A}_{100}-{A}_{90}$ is _______.
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   Answer: 18900

   Given ${l}_{1},{l}_{2}...{l}_{100}$ are consecutive terms of an $A.P$ Now let ${T}_{1}=a$ and common difference $={d}_{1}$ And similarly for $A.P$ ${w}_{1},{w}_{2},...{w}_{100}$, ${T}_{1}=b$ and common difference $={d}_{2}$ Now given, ${A}_{51}-{A}_{50}={l}_{51}{w}_{51}-{l}_{50}{w}_{50}$ $\Rightarrow \left(a+50{d}_{1}\right)\left(b+50{d}_{2}\right)-\left(a+49{d}_{1}\right)\left(b+49{d}_{2}\right)=1000$ $\Rightarrow 50b{d}_{1}+50a{d}_{2}+2500{d}_{1}{d}_{2}-49a{d}_{2}-49b{d}_{1}-2401{d}_{1}{d}_{2}=1000$ $\Rightarrow b{d}_{1}+a{d}_{2}+99{d}_{1}{d}_{2}=1000$ So, $b{d}_{1}+a{d}_{2}=10$ $\left\{\text{as given}{d}_{1}{d}_{2}=10\right\}$ Now finding ${A}_{100}-{A}_{90}={l}_{100}{w}_{100}-{l}_{90}{w}_{90}$ we get, $=\left(a+99{d}_{1}\right)\left(b+99{d}_{2}\right)-\left(a+89{d}_{1}\right)\left(b+89{d}_{2}\right)$ $=99b{d}_{1}+99a{d}_{2}+{99}^{2}{d}_{1}{d}_{2}-89b{d}_{1}-89a{d}_{2}-{89}^{2}{d}_{1}{d}_{2}$ $=10\left(b{d}_{1}+a{d}_{2}\right)+1880{d}_{1}{d}_{2}$ $=10\left(10\right)+18800$ $\left\{\text{again using}{d}_{1}{d}_{2}=10\&b{d}_{1}+a{d}_{2}=10\right\}$ $=18900$
5. Q5JEE Advanced Adv 2021 (Paper 2)
   Paragraph: Let $M=\left\{(x, y) \in \mathbb{R} \times \mathbb{R}: x^{2}+y^{2} \leq r^{2}\right\}$, where $r\gt 0 .$ Consider the geometric progression $a_{n}=\frac{1}{2^{n-1}}, n=1,2,3, \ldots .$ Let $S_{0}=0$ and, for $n \geq 1$, let $S_{n}$ denote the sum of the first $n$ terms of this progression. For $n \geq 1$, let $C_{n}$ denote the circle with center $\left(S_{n-1}, 0\right)$ and radius $a_{n}$, and $D_{n}$ denote the circle with center $\left(S_{n-1}, S_{n-1}\right)$ and radius $a_{n}$. Question: Consider $M$ with $r=\frac{\left(2^{199}-1\right) \sqrt{2}}{2^{198}} .$ The number of all those circles $D_{n}$ that are inside $M$ is
   1. A.$198$
   2. B.$199$
   3. C.$200$
   4. D.$201$
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   Answer: (B)

   $\sqrt{2}{S}_{n-1}+{a}_{n}\lt \left(\dfrac{{2}^{199}-1}{{2}^{198}}\right)\sqrt{2}$ $\sqrt{2}\left(2-\dfrac{1}{{2}^{n-2}}\right)+\dfrac{1}{{2}^{n-1}}\lt \left(\dfrac{{2}^{199}-1}{{2}^{198}}\right)\sqrt{2}$ $2\sqrt{2}-\dfrac{1}{{2}^{n-2}}\sqrt{2}+\dfrac{1}{{2}^{n-1}}\lt 2\sqrt{2}-\dfrac{\sqrt{2}}{{2}^{198}}$ $\dfrac{1}{{2}^{n-2}}\left(\dfrac{1}{2}-\sqrt{2}\right)\lt -\dfrac{\sqrt{2}}{{2}^{198}}$ $\dfrac{1}{{2}^{n-2}}\dfrac{(2\sqrt{2}-1)}{2}\gt \dfrac{\sqrt{2}}{{2}^{198}}$ ${2}^{n-2}\lt \left(2-\dfrac{1}{\sqrt{2}}\right){2}^{197}$ $n-2\leq 197$ number of circle$=199$