Sequence and Series: JEE Main 2026 PYQs

Answer: Option (2)

Key idea: Intersection of two arithmetic progressions and divisibility conditions

1. Step 1: Write general terms: Set $A$ has $a_n = 5n - 4$ for $n = 1$ to $101$, Set $B$ has $b_m = 7m + 2$ for $m = 1$ to $71$
2. Step 2: For $A \cap B$, solve $5n - 4 = 7m + 2$, which gives $5n = 7m + 6$. For integer $n$, we need $7m \equiv -1 \pmod{5}$, so $m \equiv 2 \pmod{5}$
3. Step 3: Values of $m$ are $2, 7, 12, 17, ..., 67$ (since $m \leq 71$ and corresponding $n \leq 101$). This gives 14 common elements
4. Step 4: Common elements form an A.P.: $16, 51, 86, 121, 156, 191, 226, 261, 296, 331, 366, 401, 436, 471$ with first term $16$ and common difference $35$
5. Step 5: For divisibility by 3: $16 + 35k \equiv 0 \pmod{3}$ gives $1 + 2k \equiv 0 \pmod{3}$, so $k \equiv 1 \pmod{3}$
6. Step 6: Values $k = 1, 4, 7, 10, 13$ give elements $51, 156, 261, 366, 471$ (all divisible by 3)
7. Step 7: Count the elements: There are $\boxed{5}$ elements in $A \cap B$ divisible by 3

Answer: 91

Key idea: Telescoping series and sum of squares formula

1. Given $\sum_{k=1}^{6} a_k = 6n^3$, recognize this as a telescoping sum where $a_k = 6(k^3 - (k-1)^3)$
2. Expand: $a_k = 6(3k^2 - 3k + 1) = 18k^2 - 18k + 6$
3. Find the difference: $a_{k+1} - a_k = 18(k+1)^2 - 18(k+1) + 6 - 18k^2 + 18k - 6 = 36k$
4. Substitute into the expression: $\frac{a_{k+1} - a_k}{36} = \frac{36k}{36} = k$
5. Evaluate: $\sum_{k=1}^{6} k^2 = \frac{6 \cdot 7 \cdot 13}{6} = 91$

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Answer: Option (2)

Key idea: Sum of series using standard formulas for sum of cubes and sum of odd numbers

1. Identify the general term: numerator is $1^3 + 2^3 + \cdots + n^3 = \left[\frac{n(n+1)}{2}\right]^2$ and denominator is $1 + 3 + 5 + \cdots + (2n-1) = n^2$
2. Simplify the $n$th term: $T_n = \frac{[n(n+1)/2]^2}{n^2} = \frac{n^2(n+1)^2}{4n^2} = \frac{(n+1)^2}{4}$
3. Verify with first few terms: $T_1 = 1$, $T_2 = \frac{9}{4}$, $T_3 = 4$
4. Sum of 8 terms: $S_8 = \frac{1}{4}\sum_{n=1}^{8}(n+1)^2 = \frac{1}{4}\sum_{m=2}^{9}m^2$
5. Calculate: $\sum_{m=2}^{9}m^2 = \frac{9(10)(19)}{6} - 1 = 285 - 1 = 284$
6. Therefore: $S_8 = \frac{284}{4} = 71$

Answer: Option (1)

Key idea: Arithmetic Progression - Sum formula and relationship between terms

1. Given: First term $a = \frac{10}{3}$, number of terms $n = 30$, and sum equals cube of last term
2. Express last term: $l = a + 29d = \frac{10}{3} + 29d$
3. Write sum formula: $S_{30} = \frac{n}{2}(a + l) = 15\left(\frac{10}{3} + \frac{10}{3} + 29d\right) = 100 + 435d$
4. Apply given condition: $S_{30} = l^3$, so $100 + 435d = \left(\frac{10}{3} + 29d\right)^3$
5. Test option (1): $d = \frac{5}{87}$ gives $l = \frac{10}{3} + \frac{145}{87} = \frac{435}{87} = 5$
6. Verify: $S_{30} = 100 + 435 \times \frac{5}{87} = 125$ and $l^3 = 5^3 = 125$ ✓

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Answer: 15220

Key idea: Finding nth term from sum formula and calculating sum of squares of A.P. terms

1. Given $S_n = 3n^2 + 5n$, find the general term using $a_n = S_n - S_{n-1}$
2. Calculate: $a_n = 3n^2 + 5n - 3(n-1)^2 - 5(n-1) = 6n + 2$
3. Verify: First term $a_1 = S_1 = 8$, common difference $d = 6$
4. The first 10 terms are: $8, 14, 20, 26, 32, 38, 44, 50, 56, 62$
5. Sum of squares: $\sum_{n=1}^{10}(6n+2)^2 = 36\sum n^2 + 24\sum n + 40$
6. Using formulas: $36(385) + 24(55) + 40 = 13860 + 1320 + 40 = 15220$

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Key idea: Partial Fractions and Telescoping Series

Answer: Option (3)

Key idea: Geometric Progression and Vieta's Formulas

1. Since $\alpha, \beta, \gamma, \delta$ are in G.P., let them be $a, ar, ar^2, ar^3$
2. Using Vieta's formulas: $\alpha + \beta = a(1+r) = 1$ and $\gamma + \delta = ar^2(1+r) = 4$
3. Dividing these equations: $r^2 = 4$, so $r = \pm 2$
4. For $r = 2$: $a = 1/3$ gives non-integer $p$. For $r = -2$: $a = -1$ works
5. With $a = -1, r = -2$: roots are $\alpha = -1, \beta = 2, \gamma = -4, \delta = 8$
6. Calculate $p = \alpha\beta = (-1)(2) = -2$ and $q = \gamma\delta = (-4)(8) = -32$
7. Therefore $|p + q| = |-2 - 32| = 34$

Answer: Option (3)

Key idea: Telescoping Series and Partial Fractions

1. Factor the denominator: $1 + 4k^4 = (2k^2 - 2k + 1)(2k^2 + 2k + 1)$
2. Use partial fractions to get: $\frac{k}{1+4k^4} = \frac{1}{4}\left(\frac{1}{2k^2 - 2k + 1} - \frac{1}{2k^2 + 2k + 1}\right)$
3. Observe that $2(k+1)^2 - 2(k+1) + 1 = 2k^2 + 2k + 1$, so the series telescopes
4. Sum telescopes to: $S_{10} = \frac{1}{4}\left(1 - \frac{1}{221}\right) = \frac{1}{4} \cdot \frac{220}{221} = \frac{55}{221}$
5. Verify $\gcd(55, 221) = 1$ since $55 = 5 \times 11$ and $221 = 13 \times 17$
6. Therefore $m + n = 55 + 221 = 276$

Answer: Option (4)

Key idea: Arithmetic Progression - Finding arithmetic means and calculating their average

1. The sequence is: $59, A_1, A_2, ..., A_{39}, 159$ with total $41$ terms
2. Find common difference: $d = \frac{159-59}{40} = \frac{100}{40} = 2.5$
3. Calculate the required terms: $A_{25} = 59 + 25(2.5) = 121.5$, $A_{28} = 129$, $A_{31} = 136.5$, $A_{36} = 149$
4. Find the mean: $\text{Mean} = \frac{121.5 + 129 + 136.5 + 149}{4} = \frac{536}{4} = 134$

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Answer: Option (1)

Key idea: Arithmetic Progression, Geometric Progression, and Quadratic Equations

1. Set up variables: A.P. has first term $a$ and common difference $d$; G.P. has first term $b$ and common ratio $r$
2. Use given conditions: $S_{10} = 160$ gives $2a + 9d = 32$, and $b + br = 8$ gives $b(1+r) = 8$
3. Apply the cross-relations: $a = r$ and $b = d$, so $2r + 9d = 32$ and $d(1+r) = 8$
4. From $d = \frac{8}{1+r}$, substitute into $2r + 9d = 32$ to get $2r^2 - 30r + 40 = 0$ or $r^2 - 15r + 20 = 0$
5. The two values of $r$ give two values of $b = \frac{8}{1+r}$: $b_1 = \frac{16}{17+\sqrt{145}}$ and $b_2 = \frac{16}{17-\sqrt{145}}$
6. Sum the values: $b_1 + b_2 = \frac{16(34)}{289-145} = \frac{544}{144} = \frac{34}{9}$

Answer: Option (3)

Key idea: Quadratic equations (sum and product of roots), optimization using calculus, geometric progression sum formula

1. Step 1: For the quadratic $(n^2-2n+2)x^2 - 3x + (n^2-2n+2)^2 = 0$, let $a = n^2-2n+2$. Product of roots = $a$ and sum of roots = $\frac{3}{a}$
2. Step 2: Find minimum of $a = n^2-2n+2 = (n-1)^2 + 1$. Minimum occurs at $n=1$, giving $\alpha = 1$
3. Step 3: To maximize sum of roots $\frac{3}{a}$, minimize $a$. Maximum sum is $\beta = \frac{3}{1} = 3$
4. Step 4: G.P. has first term $\alpha = 1$ and common ratio $r = \frac{\alpha}{\beta} = \frac{1}{3}$
5. Step 5: Sum of first 6 terms: $S_6 = 1 \cdot \frac{1-(1/3)^6}{1-1/3} = \frac{1-1/729}{2/3} = \frac{728/729}{2/3}$
6. Step 6: Simplify: $S_6 = \frac{728}{729} \times \frac{3}{2} = \frac{2184}{1458} = \frac{364}{243}$

Answer: Option (3)

Key idea: Sum of series using formula for sum of squares

1. Identify the general term: $T_n = \frac{1}{n}(1^2 + 2^2 + ... + n^2)$
2. Use the formula $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ to get $T_n = \frac{(n+1)(2n+1)}{6}$
3. Expand: $(n+1)(2n+1) = 2n^2 + 3n + 1$
4. Write $S_{10} = \frac{1}{6}\sum_{n=1}^{10}(2n^2 + 3n + 1) = \frac{1}{6}[2\sum n^2 + 3\sum n + 10]$
5. Calculate: $\sum_{n=1}^{10} n^2 = 385$, $\sum_{n=1}^{10} n = 55$
6. Substitute: $S_{10} = \frac{1}{6}[2(385) + 3(55) + 10] = \frac{945}{6} = \frac{315}{2}$

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Key idea: Arithmetic Progressions, Trigonometric Equations, and Trigonometric Identities