Sequences and Series JEE Main previous year questions with solutions

4 solved JEE Main questions on Sequences and Series, free to read — no sign-in needed. The full chapter has 307 questions; sign in to attempt the remaining 303 in the exam simulator.

1. Q1JEE Main 2026 (08 Apr, Shift 2)Arithmetic Progression
   Let $\alpha = 3+4+8+9+13+14+\ldots$ upto 40 terms. If $(\tan\beta)^{\frac{\alpha}{1020}}$ is a root of the equation $x^2+x-2=0$, $\beta \in \left(0, \dfrac{\pi}{2}\right)$, then $\sin^2\beta + 3\cos^2\beta$ is equal to:
   1. A.$2$
   2. B.$\dfrac{7}{4}$
   3. C.$\dfrac{5}{2}$
   4. D.$\dfrac{3}{2}$
   Show answer & solution

   Answer: (A)

   The given series is $\alpha = 3+4+8+9+13+14+\ldots$ up to $40$ terms. We can split the series into two arithmetic progressions, each containing $20$ terms: $S_1 = 3 + 8 + 13 + \ldots$ up to $20$ terms $S_2 = 4 + 9 + 14 + \ldots$ up to $20$ terms Using the sum formula for an AP, $S_n = \dfrac{n}{2}[2a + (n-1)d]$: $S_1 = \dfrac{20}{2}[2(3) + (20-1)5] = 10[6 + 95] = 1010$ $S_2 = \dfrac{20}{2}[2(4) + (20-1)5] = 10[8 + 95] = 1030$ Therefore, $\alpha = S_1 + S_2 = 1010 + 1030 = 2040$. The exponent in the given expression is $\dfrac{\alpha}{1020} = \dfrac{2040}{1020} = 2$. Thus, $(\tan\beta)^2 = \tan^2\beta$ is a root of the equation $x^2+x-2=0$. Solving $x^2+x-2=0$ gives $(x+2)(x-1)=0 \Rightarrow x = -2$ or $x = 1$. Since $\tan^2\beta \ge 0$, we must have $\tan^2\beta = 1$. Given $\beta \in \left(0, \dfrac{\pi}{2}\right)$, $\tan\beta = 1 \Rightarrow \beta = \dfrac{\pi}{4}$. We need to find the value of $\sin^2\beta + 3\cos^2\beta$. Substituting $\beta = \dfrac{\pi}{4}$: $\sin^2\left(\dfrac{\pi}{4}\right) + 3\cos^2\left(\dfrac{\pi}{4}\right) = \left(\dfrac{1}{\sqrt{2}}\right)^2 + 3\left(\dfrac{1}{\sqrt{2}}\right)^2 = \dfrac{1}{2} + \dfrac{3}{2} = 2$. Answer: $2$
2. Q2JEE Main 2025 (07 Apr, Shift 2)Geometric Progression
   If the sum of the second, fourth and sixth terms of a G.P. of positive terms is 21 and the sum of its eighth, tenth and twelfth terms is 15309 , then the sum of its first nine terms is :
   1. A.760
   2. B.755
   3. C.750
   4. D.757
   Show answer & solution

   Answer: (D)

   $\begin{aligned} & \mathrm{ar}+\mathrm{ar}^3+\mathrm{ar}^5=21, \quad \mathrm{ar}^7+\mathrm{ar}^9+\mathrm{ar}^{11}=15309 \\ & \Rightarrow \operatorname{ar}\left(1+\mathrm{r}^2+\mathrm{r}^4\right)=21, \quad \operatorname{ar}^7\left(1+\mathrm{r}^2+\mathrm{r}^4\right)=15309\end{aligned}$ $\mathrm{Eq}^{\mathrm{n}}\left({\overline{2)} \div \mathrm{eq}^{\mathrm{n}}(1) \quad \quad \quad \text { (2) }}_{(1)}^{(2)}\right.$ $\begin{aligned} & \Rightarrow \frac{\mathrm{a} \cdot \mathrm{r}^7}{\mathrm{ar}}=\frac{15309}{21} \Rightarrow \mathrm{r}^6=729 \\ & \Rightarrow \frac{\mathrm{a} \cdot\left(\mathrm{r}^9-1\right)}{\mathrm{r}-1}=\frac{\frac{7}{91}(19683-1)}{2}=\frac{7 \times 19682}{91 \times 2} \\ & =\frac{9841}{13}=757\end{aligned}$
3. Q3JEE Main 2024 (04 Apr, Shift 2)Mixed Question on AP and GP
   Let three real numbers $a, b, c$ be in arithmetic progression and $a+1, b, c+3$ be in geometric progression. If $a\gt 10$ and the arithmetic mean of $a, b$ and $c$ is 8, then the cube of the geometric mean of $a, b$ and $c$ is
   1. A.128
   2. B.316
   3. C.120
   4. D.312
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   Answer: (C)

   $\begin{aligned} & 2 b=a+c, b^2=(a+1)(c+3) \\ & \frac{a+b+c}{3}=8 \rightarrow b=8, a+c=16 \\ & 64=(a+1)(19-a)=19+18 a-a^2 \\ & a^2-18 a-45=0 \rightarrow(a-15)(a+3)=0,(a\gt 10) \\ & a=15, c=1, b=8 \\ & \left((a b c)^{1 / 3}\right)^3=a b c=120\end{aligned}$
4. Q4JEE Main 2023 (11 Apr, Shift 2)Inequalities
   Let $a,b,c$ and $d$ be positive real numbers such that $a+b+c+d=11$. If the maximum value of ${a}^{5}{b}^{3}{c}^{2}d$ is $3750\beta$, then the value of $\beta$ is
   1. A.$90$
   2. B.$110$
   3. C.$55$
   4. D.$108$
   Show answer & solution

   Answer: (A)

   Given: $a+b+c+d=11$ We know that, $AM\geq GM$ $\Rightarrow \dfrac{\dfrac{a}{5}+\dfrac{a}{5}+\dfrac{a}{5}+\dfrac{a}{5}+\dfrac{a}{5}+\dfrac{b}{3}+\dfrac{b}{3}+\dfrac{b}{3}+\dfrac{c}{2}+\dfrac{c}{2}+\dfrac{d}{1}}{11}\geq \sqrt[11]{\dfrac{{a}^{5}{b}^{3}{c}^{2}d}{{5}^{5}{3}^{3}{2}^{2}}}$ $\Rightarrow \dfrac{a+b+c+d}{11}\geq \sqrt[11]{\dfrac{{a}^{5}{b}^{3}{c}^{2}d}{{5}^{5}{3}^{3}{2}^{2}}}$ $\Rightarrow \dfrac{11}{11}\geq \sqrt[11]{\dfrac{{a}^{5}{b}^{3}{c}^{2}d}{{5}^{5}{3}^{3}{2}^{2}}}$ $\Rightarrow \sqrt[11]{\dfrac{{a}^{5}{b}^{3}{c}^{2}d}{{5}^{5}{3}^{3}{2}^{2}}}\leq 1$ $\Rightarrow {a}^{5}{b}^{3}{c}^{2}d\leq {5}^{5}{3}^{3}{2}^{2}$ $\Rightarrow {a}^{5}{b}^{3}{c}^{2}d\leq 90\cdot 3750$ So, $\beta =90$