Sequences and Series Previous Year Questions (JEE Main 2026)

Let $a = \dfrac{10}{3}$, $n = 30$, common difference $d$, and last term $\ell = a + 29d$.

The sum is

$$S = \frac{n}{2}(a + \ell) = 15(a + \ell).$$

The condition $S = \ell^3$ gives

$$15\left(\frac{10}{3} + \ell\right) = \ell^3 \;\Rightarrow\; 50 + 15\ell = \ell^3.$$

So $\ell^3 - 15\ell - 50 = 0$. Testing $\ell = 5$: $125 - 75 - 50 = 0$. Hence $\ell = 5$.

Then

$$a + 29d = 5 \;\Rightarrow\; \frac{10}{3} + 29d = 5 \;\Rightarrow\; 29d = \frac{5}{3} \;\Rightarrow\; d = \frac{5}{87}.$$

All terms $a + kd$ for $k = 0,\dots,29$ are non-negative, so the condition holds.

Answer: A ($\frac{5}{87}$)

Group the 15 terms as $(1^3 - 2^3) + (3^3 - 4^3) + \dots + (13^3 - 14^3) + 15^3$.

Each odd cube minus the next even cube uses $a^3 - b^3 = (a-b)(a^2+ab+b^2)$ with $a - b = -1$, but it is cleaner to compute directly. Split into odd and even cubes:

$$\underbrace{(1^3 + 3^3 + \dots + 15^3)}_{\text{odd}} - \underbrace{(2^3 + 4^3 + \dots + 14^3)}_{\text{even}}.$$

Sum of all cubes to $15$: $\left(\dfrac{15\cdot 16}{2}\right)^2 = 120^2 = 14400.$

Even cubes: $2^3(1^3 + \dots + 7^3) = 8\left(\dfrac{7\cdot 8}{2}\right)^2 = 8\cdot 784 = 6272.$

Odd cubes $= 14400 - 6272 = 8128$. Therefore the alternating sum is

$$8128 - 6272 = 1856.$$

Answer: B (1856)

Let the A.P. have first term $A$ and common difference $D$. Let the G.P. have first term $g$ and common ratio $r$.

Given links: $A = r$ and $g = D$.

Sum of first ten A.P. terms:

$$\frac{10}{2}(2A + 9D) = 160 \;\Rightarrow\; 2A + 9D = 32.$$

Sum of first two G.P. terms with $g = D$, $r = A$:

$$g + gr = D(1 + A) = 8.$$

From the first equation $A = \dfrac{32 - 9D}{2}$. Substituting:

$$D\left(1 + \frac{32 - 9D}{2}\right) = 8 \;\Rightarrow\; D(34 - 9D) = 16 \;\Rightarrow\; 9D^2 - 34D + 16 = 0.$$

The first term of the G.P. is $g = D$, so the sum of all possible values of $g$ is the sum of the roots of this quadratic:

$$D_1 + D_2 = \frac{34}{9}.$$

Answer: A ($\dfrac{34}{9}$)

For $f(\theta) = \alpha\tan^2\theta + \beta\cot^2\theta$, by AM–GM

$$f(\theta) \ge 2\sqrt{\alpha\beta}, \qquad \min f = 2\sqrt{\alpha\beta}.$$

For $g(\theta) = \alpha\sin^2\theta + \beta\cos^2\theta = \beta + (\alpha - \beta)\sin^2\theta$ with $\alpha > \beta$, the maximum over $0 < \theta < \pi$ occurs at $\sin^2\theta = 1$:

$$\max g = \alpha.$$

Setting them equal:

$$2\sqrt{\alpha\beta} = \alpha \;\Rightarrow\; 4\alpha\beta = \alpha^2 \;\Rightarrow\; \alpha = 4\beta.$$

Then first term $= \dfrac{\alpha}{2\beta} = 2$ and ratio $= \dfrac{2\beta}{\alpha} = \dfrac{1}{2}$.

$$S_{10} = 2\cdot\frac{1 - (1/2)^{10}}{1 - 1/2} = 4\left(1 - \frac{1}{1024}\right) = \frac{1023}{256}.$$

Here $m = 1023$, $n = 256$, so $m + n = 1279$.

Answer: 1279

The general term is $(x + k)(x + k + 2)$ for $k = 0, 1, \dots, n-1$, i.e. $(x+k)^2 + 2(x+k)$.

Summing:

$$\sum_{k=0}^{n-1}\big[(x+k)^2 + 2(x+k)\big] = nx^2 + 2x\sum k + \sum k^2 + 2nx + 2\sum k.$$

With $\sum_{k=0}^{n-1} k = \dfrac{n(n-1)}{2}$ and $\sum_{k=0}^{n-1} k^2 = \dfrac{(n-1)n(2n-1)}{6}$, the equation $= 4n$ becomes a quadratic $nx^2 + bx + c = 0$ whose roots differ by $2$.

For roots $\alpha$ and $\alpha + 2$, the difference of roots is $2$, so

$$(\text{root difference})^2 = \frac{b^2 - 4nc'}{n^2} = 4,$$

where $c' = c - 4n$. Solving over $n \in \mathbb{N}$ with integer $\alpha$, the only case that works is $n = 7$, giving roots $-5$ and $-3$, so $\alpha = -5$.

(Check $n = 7$: the quadratic is $7x^2 + 56x + 105 = 0 \Rightarrow x^2 + 8x + 15 = 0 \Rightarrow x = -3, -5$.)

Therefore $n + \alpha = 7 + (-5) = 2$.

Answer: C (2)

From the partial sum, for $k \ge 1$

$$a_k = S_k - S_{k-1} = 6k^3 - 6(k-1)^3 = 6\big(3k^2 - 3k + 1\big) = 18k^2 - 18k + 6.$$

Then

$$a_{k+1} - a_k = 18\big[(k+1)^2 - k^2\big] - 18\big[(k+1) - k\big] = 18(2k+1) - 18 = 36k.$$

So $\dfrac{a_{k+1} - a_k}{36} = k$, and

$$\sum_{k=1}^{6} k^2 = \frac{6\cdot 7\cdot 13}{6} = 91.$$

Answer: 91

Both are infinite geometric series:

$$\alpha = \frac{1/4}{1 - 1/2} = \frac{1}{2}, \qquad \beta = \frac{1/3}{1 - 1/3} = \frac{1}{2}.$$

First term: $0.2 = 5^{-1}$ and $\log_{\sqrt5}(1/2) = 2\log_5(1/2)$, so

$$(0.2)^{\log_{\sqrt5}(1/2)} = 5^{-2\log_5(1/2)} = \left(\tfrac{1}{2}\right)^{-2} = 4.$$

Second term: $0.04 = 5^{-2}$, so

$$(0.04)^{\log_5(1/2)} = 5^{-2\log_5(1/2)} = \left(\tfrac{1}{2}\right)^{-2} = 4.$$

Sum $= 4 + 4 = 8$.

Answer: C (8)

Let the A.P. have first term $a_1 = 1$ and common difference $d$, and the G.P. have $g_1 = 1$ and ratio $r$ (with $r > 1$ since it is increasing).

From $a_2 + g_2 = 1$:

$$(1 + d) + r = 1 \;\Rightarrow\; d = -r.$$

From $a_3 + g_3 = 4$:

$$(1 + 2d) + r^2 = 4 \;\Rightarrow\; 1 - 2r + r^2 = 4 \;\Rightarrow\; r^2 - 2r - 3 = 0.$$

So $(r - 3)(r + 1) = 0$, giving $r = 3$ (taking the increasing root). Then $d = -3$.

$$a_{10} = 1 + 9d = 1 - 27 = -26, \qquad g_5 = r^4 = 81.$$$$a_{10} + g_5 = -26 + 81 = 55.$$

Answer: D ($55$)

The $r$-th term has numerator $1^3 + \dots + r^3 = \left(\dfrac{r(r+1)}{2}\right)^2$ and denominator $1 + 3 + \dots + (2r-1) = r^2$.

So

$$T_r = \frac{\left(\frac{r(r+1)}{2}\right)^2}{r^2} = \frac{(r+1)^2}{4}.$$

Summing to $8$ terms:

$$\sum_{r=1}^{8} \frac{(r+1)^2}{4} = \frac{1}{4}\big(2^2 + 3^2 + \dots + 9^2\big) = \frac{1}{4}\big(285 - 1\big) = \frac{284}{4} = 71.$$

(Here $\sum_{k=1}^{9} k^2 = 285$, minus $1^2$.)

Answer: B ($71$)

Let $t = n^2 - 2n + 2 = (n-1)^2 + 1 \ge 1$.

By Vieta’s formulas:

$$\text{product of roots} = \frac{t^2}{t} = t, \qquad \text{sum of roots} = \frac{3}{t}.$$

• Minimum product $= \min t = 1$ (at $n = 1$), so $\alpha = 1$.
• Maximum sum $= \max \dfrac{3}{t} = \dfrac{3}{1} = 3$, so $\beta = 3$.

G.P. first term $\alpha = 1$, ratio $\dfrac{\alpha}{\beta} = \dfrac{1}{3}$:

$$S_6 = \frac{1 - (1/3)^6}{1 - 1/3} = \frac{3}{2}\left(1 - \frac{1}{729}\right) = \frac{3}{2}\cdot\frac{728}{729} = \frac{364}{243}.$$

Answer: C ($\dfrac{364}{243}$)

The $r$-th term is

$$T_r = \frac{1}{r}\sum_{i=1}^{r} i^2 = \frac{1}{r}\cdot\frac{r(r+1)(2r+1)}{6} = \frac{(r+1)(2r+1)}{6} = \frac{2r^2 + 3r + 1}{6}.$$

Summing to $10$ terms:

$$\sum_{r=1}^{10} T_r = \frac{1}{6}\left(2\sum r^2 + 3\sum r + \sum 1\right) = \frac{1}{6}\big(2\cdot 385 + 3\cdot 55 + 10\big).$$$$= \frac{1}{6}(770 + 165 + 10) = \frac{945}{6} = \frac{315}{2}.$$

Answer: C ($\dfrac{315}{2}$)

The $n$-th term is

$$a_n = S_n - S_{n-1} = (3n^2 + 5n) - \big(3(n-1)^2 + 5(n-1)\big) = 6n + 2.$$

So the terms are $8, 14, 20, \dots$ up to $a_{10} = 62$.

$$\sum_{n=1}^{10} a_n^2 = \sum_{n=1}^{10} (6n+2)^2 = \sum_{n=1}^{10}\big(36n^2 + 24n + 4\big).$$$$= 36\cdot 385 + 24\cdot 55 + 4\cdot 10 = 13860 + 1320 + 40 = 15220.$$

Answer: C ($15220$)

Use the telescoping identity

$$\frac{1}{n(n+1)(n+2)} = \frac{1}{2}\left[\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)}\right].$$

Therefore

$$\sum_{n=1}^{10}\frac{1}{n(n+1)(n+2)} = \frac{1}{2}\left[\frac{1}{1\cdot 2} - \frac{1}{11\cdot 12}\right] = \frac{1}{2}\left[\frac{1}{2} - \frac{1}{132}\right].$$$$= \frac{1}{2}\cdot\frac{66 - 1}{132} = \frac{65}{264}.$$

Multiplying by $528$:

$$528\cdot\frac{65}{264} = 2\cdot 65 = 130.$$

Answer: B ($130$)

Vieta: $\alpha + \beta = 1$, $\alpha\beta = p$; $\gamma + \delta = 4$, $\gamma\delta = q$.

Let the G.P. be $A, Ar, Ar^2, Ar^3$ so $\alpha = A$, $\beta = Ar$, $\gamma = Ar^2$, $\delta = Ar^3$.

$$\alpha + \beta = A(1 + r) = 1, \qquad \gamma + \delta = Ar^2(1 + r) = 4.$$

Dividing: $r^2 = 4 \Rightarrow r = \pm 2$.

For $r = 2$: $A = \frac{1}{3}$, giving $p = \alpha\beta = \frac{2}{9} \notin \mathbb{Z}$ — rejected.

For $r = -2$: $A(1 - 2) = 1 \Rightarrow A = -1$. Then

$$\alpha = -1,\; \beta = 2,\; \gamma = -4,\; \delta = 8.$$$$p = \alpha\beta = -2, \qquad q = \gamma\delta = -32 \quad (\text{both integers}).$$$$|p + q| = |-2 - 32| = 34.$$

Answer: C (34)

The general term is $\dfrac{r}{1 + 4r^4}$. Factor the denominator (Sophie Germain identity):

$$1 + 4r^4 = (2r^2 - 2r + 1)(2r^2 + 2r + 1).$$

Note that $(2r^2 + 2r + 1) - (2r^2 - 2r + 1) = 4r$, so

$$\frac{r}{1 + 4r^4} = \frac{1}{4}\left[\frac{1}{2r^2 - 2r + 1} - \frac{1}{2r^2 + 2r + 1}\right].$$

This telescopes. With $f(r) = 2r^2 - 2r + 1$, note $2r^2 + 2r + 1 = f(r+1)$, so

$$\sum_{r=1}^{10} = \frac{1}{4}\left[\frac{1}{f(1)} - \frac{1}{f(11)}\right] = \frac{1}{4}\left[1 - \frac{1}{221}\right] = \frac{1}{4}\cdot\frac{220}{221} = \frac{55}{221}.$$

Here $\gcd(55, 221) = 1$, so $m + n = 55 + 221 = 276$.

Answer: C (276)

Inserting $39$ arithmetic means between $59$ and $159$ gives an A.P. of $41$ terms:

$$59, A_1, A_2, \dots, A_{39}, 159.$$

Common difference:

$$d = \frac{159 - 59}{40} = \frac{100}{40} = \frac{5}{2}.$$

Then $A_k = 59 + k\cdot\dfrac{5}{2}$. The required mean is

$$\frac{A_{25} + A_{28} + A_{31} + A_{36}}{4} = 59 + \frac{5}{2}\cdot\frac{25 + 28 + 31 + 36}{4} = 59 + \frac{5}{2}\cdot\frac{120}{4}.$$$$= 59 + \frac{5}{2}\cdot 30 = 59 + 75 = 134.$$

Answer: D (134)

The terms come in pairs: $(3,4), (8,9), (13,14), \dots$ The first elements $3, 8, 13, \dots$ form an A.P. with difference $5$, and the second element of each pair is one more than the first.

40 terms $=$ 20 pairs. The $i$-th pair is $(5i - 2,\, 5i - 1)$ with pair sum $10i - 3$:

$$\alpha = \sum_{i=1}^{20}(10i - 3) = 10\cdot\frac{20\cdot 21}{2} - 3\cdot 20 = 2100 - 60 = 2040.$$

So $\dfrac{\alpha}{1020} = 2$.

The roots of $x^2 + x - 2 = 0$ are $x = 1$ and $x = -2$. Since $\beta \in (0, \frac{\pi}{2})$, $\tan\beta > 0$, so $(\tan\beta)^2 > 0$, forcing the root to be $1$:

$$(\tan\beta)^2 = 1 \;\Rightarrow\; \tan\beta = 1 \;\Rightarrow\; \beta = \frac{\pi}{4}.$$

Then $\sin^2\beta = \cos^2\beta = \frac{1}{2}$, so

$$\sin^2\beta + 3\cos^2\beta = \frac{1}{2} + \frac{3}{2} = 2.$$

Answer: A (2)

The infinite series is geometric:

$$2 + \frac{2}{3} + \frac{2}{9} + \ldots = \frac{2}{1 - 1/3} = 3.$$

So the relation becomes

$$\log_2 f(x) = \log_2 3 \cdot \log_3\!\left(1 + \frac{f(x)}{f(1/x)}\right) = \log_2\!\left(1 + \frac{f(x)}{f(1/x)}\right),$$

using $\log_2 3 \cdot \log_3 y = \log_2 y$. Hence

$$f(x) = 1 + \frac{f(x)}{f(1/x)}.$$

Try $f(x) = 1 + x^k$. Then

$$\frac{f(x)}{f(1/x)} = \frac{1 + x^k}{1 + x^{-k}} = \frac{1 + x^k}{(x^k + 1)/x^k} = x^k,$$

so $1 + \frac{f(x)}{f(1/x)} = 1 + x^k = f(x)$. Consistent.

Using $f(6) = 37$:

$$1 + 6^k = 37 \;\Rightarrow\; 6^k = 36 \;\Rightarrow\; k = 2, \quad f(x) = 1 + x^2.$$

Therefore

$$\sum_{n=1}^{10} f(n) = \sum_{n=1}^{10}(1 + n^2) = 10 + \frac{10\cdot 11\cdot 21}{6} = 10 + 385 = 395.$$

Answer: 395