Solution & Explanation
### Related Formula
For series whose consecutive differences form an Arithmetic Progression (A.P.), the general term is given by a quadratic form:
T_n = an^2 + bn + c$T_n = an^2 + bn + c$
### Core Logic
Analyze successive first-order differences of the terms [cite: 1293, 1294]:
textSeries: 1, quad 3, quad 11, quad 25, quad 45, quad 71$\text{Series: } 1, \quad 3, \quad 11, \quad 25, \quad 45, \quad 71$ [cite: 1293]
textDifferences: 2, quad 8, quad 14, quad 20, quad 26$\text{Differences: } 2, \quad 8, \quad 14, \quad 20, \quad 26$ [cite: 1294]
Since the differences grow uniformly by 6, they reside in an A.P. [cite: 1294]
Thus, set up the general term system [cite: 1296, 1298]:
- T_1 = a + b + c = 1$T_1 = a + b + c = 1$
- T_2 = 4a + 2b + c = 3$T_2 = 4a + 2b + c = 3$
- T_3 = 9a + 3b + c = 11$T_3 = 9a + 3b + c = 11$
Solving the linear equations simultaneously yields [cite: 1299]:
a = 3, quad b = -7, quad c = 5$a = 3, \quad b = -7, \quad c = 5$ [cite: 1299]
### Step 1: Summing the Series
The general term is [cite: 1300]:
T_n = 3n^2 - 7n + 5$T_n = 3n^2 - 7n + 5$ [cite: 1300]
Evaluate the summation for n=20$n=20$ terms [cite: 1302]:
S_20 = sum_n=1^20 (3n^2 - 7n + 5) = 3sum_n=1^20 n^2 - 7sum_n=1^20 n + sum_n=1^20 5$S_{20} = \sum_{n=1}^{20} (3n^2 - 7n + 5) = 3\sum_{n=1}^{20} n^2 - 7\sum_{n=1}^{20} n + \sum_{n=1}^{20} 5$ [cite: 1302]
Substitute standard sequence formulas [cite: 1302]:
S_20 = 3 cdot left(frac20 cdot 21 cdot 416right) - 7 cdot left(frac20 cdot 212right) + 5(20)$S_{20} = 3 \cdot \left(\frac{20 \cdot 21 \cdot 41}{6}\right) - 7 \cdot \left(\frac{20 \cdot 21}{2}\right) + 5(20)$ [cite: 1302]
= 8610 - 1470 + 100 = 7240$= 8610 - 1470 + 100 = 7240$ [cite: 1302]
### Pattern Recognition
When first-order differences form a regular arithmetic line, the original function is exactly quadratic. Identify coefficients using small terms quickly.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Mathematics: Sequences and Series
More Sequences and Series Previous-Year Questions — Page 4
Q70
2025
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 :
- A. 745$745$
- B. 755$755$
- C. 750$750$
- D. 757$757$
Solution
### Related Formula
Sum of first n$n$ terms of a Geometric Progression (GP) is:
S_n = fraca(r^n - 1)r - 1$S_n = \frac{a(r^n - 1)}{r - 1}$
### Core Logic
Let the first term be a$a$ and common ratio be r$r$.
Given:
1) ar + ar^3 + ar^5 = 21 implies ar(1 + r^2 + r^4) = 21 quad dots text(1)$ar + ar^3 + ar^5 = 21 \implies ar(1 + r^2 + r^4) = 21 \quad \dots \text{(1)}$
2) ar^7 + ar^9 + ar^11 = 15309 implies ar^7(1 + r^2 + r^4) = 15309 quad dots text(2)$ar^7 + ar^9 + ar^{11} = 15309 \implies ar^7(1 + r^2 + r^4) = 15309 \quad \dots \text{(2)}$
Dividing equation (2) by equation (1):
fracar^7ar = frac1530921 implies r^6 = 729 implies r = 3$\frac{ar^7}{ar} = \frac{15309}{21} \implies r^6 = 729 \implies r = 3$
### Step 1: Solve for a
Substitute r = 3$r = 3$ into equation (1):
a(3)(1 + 9 + 81) = 21$a(3)(1 + 9 + 81) = 21$
3a(91) = 21 implies a = frac791 = frac113$3a(91) = 21 \implies a = \frac{7}{91} = \frac{1}{13}$
### Step 2: Find Sum of 9 terms
Evaluating S_9$S_9$:
S_9 = fraca(r^9 - 1)r - 1 = fracfrac113(3^9 - 1)3 - 1 = frac19683 - 126 = frac1968226 = 757$S_9 = \frac{a(r^9 - 1)}{r - 1} = \frac{\frac{1}{13}(3^9 - 1)}{3 - 1} = \frac{19683 - 1}{26} = \frac{19682}{26} = 757$
### Pattern Recognition
Ratios of shifted groups of terms in a GP always cleanly isolate a simple power of the common ratio r^k$r^k$ instantly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Sequences and Series
Q59
2025
Arithmetic Progression Sum
In an arithmetic progression, if S_40=1030$S_{40}=1030$ and S_12=57,$S_{12}=57,$ then S_30-S_10$S_{30}-S_{10}$ is equal to: [cite: 3294, 3295, 3296]
- A. 510$510$
- B. 515$515$
- C. 525$525$
- D. 505$505$
Solution
### Related Formula
Sum of first n$n$ terms of an Arithmetic Progression:
S_n = fracn2[2a + (n-1)d]$S_n = \frac{n}{2}[2a + (n-1)d]$
### Core Logic
Set up linear expressions for the given sums :
S_40 = frac402[2a + 39d] = 1030 Rightarrow 2a + 39d = 51.5$S_{40} = \frac{40}{2}[2a + 39d] = 1030 \Rightarrow 2a + 39d = 51.5$
S_12 = frac122[2a + 11d] = 57 Rightarrow 2a + 11d = 9.5$S_{12} = \frac{12}{2}[2a + 11d] = 57 \Rightarrow 2a + 11d = 9.5$
### Step 1: Solve for a$a$ and d$d$
Subtract the second equation from the first :
(2a + 39d) - (2a + 11d) = 51.5 - 9.5$(2a + 39d) - (2a + 11d) = 51.5 - 9.5$
28d = 42 Rightarrow d = frac4228 = frac32 = 1.5$28d = 42 \Rightarrow d = \frac{42}{28} = \frac{3}{2} = 1.5$
Substitute d = 1.5$d = 1.5$ back to find a$a$:
2a + 11(1.5) = 9.5 Rightarrow 2a + 16.5 = 9.5 Rightarrow 2a = -7 Rightarrow a = -3.5$2a + 11(1.5) = 9.5 \Rightarrow 2a + 16.5 = 9.5 \Rightarrow 2a = -7 \Rightarrow a = -3.5$
### Step 2: Evaluate S_30 - S_10$S_{30} - S_{10}$
Write out the formula for the target subtraction :
S_30 - S_10 = frac302[2a + 29d] - frac102[2a + 9d]$S_{30} - S_{10} = \frac{30}{2}[2a + 29d] - \frac{10}{2}[2a + 9d]$
= 15(2a + 29d) - 5(2a + 9d) = 30a + 435d - 10a - 45d = 20a + 390d$= 15(2a + 29d) - 5(2a + 9d) = 30a + 435d - 10a - 45d = 20a + 390d$ [cite: 3964, 3965]
Substitute the values of a$a$ and d$d$ :
= 20(-3.5) + 390(1.5) = -70 + 585 = 515$= 20(-3.5) + 390(1.5) = -70 + 585 = 515$
### Pattern Recognition
Notice that S_30 - S_10$S_{30} - S_{10}$ represents the \sum of terms from T_11$T_{11}$ to T_30$T_{30}$, which can also be formulated as 20 times A_20.5$20 \times A_{20.5}$, saving algebraic steps if calculated symmetrically.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Sequences and Series
Q60
2025
Arithmetico-Geometric Progression
If 7=5+frac17(5+alpha)+frac17^2(5+2alpha)+frac17^3(5+3alpha)+dotsdotsinfty$7=5+\frac{1}{7}(5+\alpha)+\frac{1}{7^{2}}(5+2\alpha)+\frac{1}{7^{3}}(5+3\alpha)+\dots\dots\infty$, then the value of alpha$\alpha$ is: [cite: 3301, 3302]
- A. 1$1$
- B. frac67$\frac{6}{7}$
- C. 6$6$
- D. frac17$\frac{1}{7}$
Solution
### Related Formula
Sum of an infinite geometric progression:
S_infty = fraca1-r quad textfor |r| < 1$S_{\infty} = \frac{a}{1-r} \quad \text{for } |r| < 1$
### Core Logic
The given expression is an infinite Arithmetico-Geometric Progression (AGP) :
S = 5 + frac5+alpha7 + frac5+2alpha7^2 + frac5+3alpha7^3 + dots infty$S = 5 + \frac{5+\alpha}{7} + \frac{5+2\alpha}{7^2} + \frac{5+3\alpha}{7^3} + \dots \infty$
### Step 1: Shift and Subtract
Multiply the equation by the common ratio frac17$\frac{1}{7}$ and shift it by one position :
frac17S = frac57 + frac5+alpha7^2 + frac5+2alpha7^3 + dots infty$\frac{1}{7}S = \frac{5}{7} + \frac{5+\alpha}{7^2} + \frac{5+2\alpha}{7^3} + \dots \infty$
Subtract this from the original equation:
S - frac17S = 5 + left(frac5+alpha-57right) + left(frac5+2alpha-(5+alpha)7^2right) + dots$S - \frac{1}{7}S = 5 + \left(\frac{5+\alpha-5}{7}\right) + \left(\frac{5+2\alpha-(5+\alpha)}{7^2}\right) + \dots$
frac67S = 5 + fracalpha7 + fracalpha7^2 + fracalpha7^3 + dots$\frac{6}{7}S = 5 + \frac{\alpha}{7} + \frac{\alpha}{7^2} + \frac{\alpha}{7^3} + \dots$
### Step 2: Sum the Infinite Geometric Series
Apply the infinite GP formula to the terms involving alpha$\alpha$ :
frac67S = 5 + fracalpha7left(frac11 - frac17right) = 5 + fracalpha7left(frac76right) = 5 + fracalpha6$\frac{6}{7}S = 5 + \frac{\alpha}{7}\left(\frac{1}{1 - \frac{1}{7}}\right) = 5 + \frac{\alpha}{7}\left(\frac{7}{6}\right) = 5 + \frac{\alpha}{6}$
Given that S = 7$S = 7$ :
frac67(7) = 5 + fracalpha6 Rightarrow 6 = 5 + fracalpha6$\frac{6}{7}(7) = 5 + \frac{\alpha}{6} \Rightarrow 6 = 5 + \frac{\alpha}{6}$
1 = fracalpha6 Rightarrow alpha = 6$1 = \frac{\alpha}{6} \Rightarrow \alpha = 6$
### Pattern Recognition
Standard trick for infinite AGPs: Multiply by the common ratio r$r$, shift, and subtract to condense the arithmetic progression component into a straightforward infinite geometric progression.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Sequences and Series
Q54
2025
Sum to n terms of Special Series
Let S_n = frac12 + frac16 + frac112 + frac120 + dots$S_{n} = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \dots$ up to n$n$ terms. If the sum of the first six terms of an A.P. with first term -p$-p$ and common difference p$p$ is sqrt2026S_2025$\sqrt{2026S_{2025}}$, then the absolute difference between 20^textth$20^{\text{th}}$ and 15^textth$15^{\text{th}}$ terms of the A.P. is :
- A. 25$25$
- B. 90$90$
- C. 20$20$
- D. 45$45$
Solution
### Related Formula
The general term for the provided series is:
T_k = frac1k(k+1) = frac1k - frac1k+1$T_k = \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$
This sets up a standard telescoping summation sequence.
### Core Logic
Express the sum S_2025$S_{2025}$ via telescoping fractions:
S_2025 = sum_k=1^2025 left( frac1k - frac1k+1 right) = left(1 - frac12right) + left(frac12 - frac13right) + dots + left(frac12025 - frac12026right)$S_{2025} = \sum_{k=1}^{2025} \left( \frac{1}{k} - \frac{1}{k+1} \right) = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \dots + \left(\frac{1}{2025} - \frac{1}{2026}\right)$
S_2025 = 1 - frac12026 = frac20252026$S_{2025} = 1 - \frac{1}{2026} = \frac{2025}{2026}$
### Step 1: Compute the boundary expression value
Substitute S_2025$S_{2025}$ into the expression value:
sqrt2026 cdot S_2025 = sqrt2026 cdot frac20252026 = sqrt2025 = 45$\sqrt{2026 \cdot S_{2025}} = \sqrt{2026 \cdot \frac{2025}{2026}} = \sqrt{2025} = 45$
### Step 2: Apply Arithmetic Progression Summation
The sum of the first 6 terms of the A.P. with a = -p$a = -p$ and d = p$d = p$ is equal to 45:
Sigma_6 = frac62 [2a + (6-1)d] = 45$\Sigma_6 = \frac{6}{2} [2a + (6-1)d] = 45$
3 [2(-p) + 5p] = 45$3 [2(-p) + 5p] = 45$
3 [3p] = 45 implies 9p = 45 implies p = 5$3 [3p] = 45 \implies 9p = 45 \implies p = 5$
### Step 3: Calculate target absolute term difference
The absolute difference between the 20^textth$20^{\text{th}}$ and 15^textth$15^{\text{th}}$ terms of any A.P. depends strictly on the common difference:
|A_20 - A_15| = |(a + 19p) - (a + 14p)| = 5p$|A_{20} - A_{15}| = |(a + 19p) - (a + 14p)| = 5p$
5p = 5(5) = 25$5p = 5(5) = 25$
### Pattern Recognition
The series sequence frac12 + frac16 + frac112 + dots$\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \dots$ is the well-known telescoping series sum frac1n(n+1)$\sum \frac{1}{n(n+1)}$. Its sum to n$n$ terms is identically given by fracnn+1$\frac{n}{n+1}$ without requiring manual re-derivation.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Sequences and Series
Q63
2025
Telescopic Series Summation
For positive integers n, if 4a_n=(n^2+5n+6)$4a_{n}=(n^{2}+5n+6)$ and S_n=sum_k=1^nleft(frac1a_kright)$S_{n}=\sum_{k=1}^{n}\left(\frac{1}{a_{k}}\right)$ then the value of 507 S_2025$S_{2025}$ is:
- A. 540$540$
- B. 1350$1350$
- C. 675$675$
- D. 135$135$
Solution
### Related Formula
Telescopic series decomposition via method of differences:
frac1(k+2)(k+3) = frac1k+2 - frac1k+3$\frac{1}{(k+2)(k+3)} = \frac{1}{k+2} - \frac{1}{k+3}$
### Core Logic
Given:
a_n = fracn^2+5n+64 = frac(n+2)(n+3)4$a_n = \frac{n^2+5n+6}{4} = \frac{(n+2)(n+3)}{4}$
Therefore, the reciprocal term is:
frac1a_k = frac4(k+2)(k+3) = 4 left[ frac1k+2 - frac1k+3 right]$\frac{1}{a_k} = \frac{4}{(k+2)(k+3)} = 4 \left[ \frac{1}{k+2} - \frac{1}{k+3} \right]$
### Step 1: Compute the Partial Sum
S_n = sum_k=1^n frac1a_k = 4 sum_k=1^n left( frac1k+2 - frac1k+3 right)$S_n = \sum_{k=1}^{n} \frac{1}{a_k} = 4 \sum_{k=1}^{n} \left( \frac{1}{k+2} - \frac{1}{k+3} \right)$
Expanding the sum terms:
S_n = 4 left[ left(frac13 - frac14right) + left(frac14 - frac15right) + dots + left(frac1n+2 - frac1n+3right) right]$S_n = 4 \left[ \left(\frac{1}{3} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{5}\right) + \dots + \left(\frac{1}{n+2} - \frac{1}{n+3}\right) \right]$
All intermediate terms cancel out:
S_n = 4 left[ frac13 - frac1n+3 right] = 4 left[ fracn+3 - 33(n+3) right] = frac4n3(n+3)$S_n = 4 \left[ \frac{1}{3} - \frac{1}{n+3} \right] = 4 \left[ \frac{n+3 - 3}{3(n+3)} \right] = \frac{4n}{3(n+3)}$
### Step 2: Calculate for n = 2025
For n = 2025$n = 2025$:
S_2025 = frac4 times 20253 times (2025 + 3) = frac4 times 20253 times 2028$S_{2025} = \frac{4 \times 2025}{3 \times (2025 + 3)} = \frac{4 \times 2025}{3 \times 2028}$
We need to find 507 times S_2025$507 \times S_{2025}$:
507 times S_2025 = 507 times frac4 times 20253 times 2028$507 \times S_{2025} = 507 \times \frac{4 \times 2025}{3 \times 2028}$
Notice that 2028 = 4 times 507$2028 = 4 \times 507$:
507 times S_2025 = 507 times frac4 times 20253 times (4 times 507) = frac20253 = 675$507 \times S_{2025} = 507 \times \frac{4 \times 2025}{3 \times (4 \times 507)} = \frac{2025}{3} = 675$
### Pattern Recognition
Always look for arithmetic factor groupings at the end of large number sequence questions in JEE. Here recognizing 2028 = 4 times 507$2028 = 4 \times 507$ avoids large multi-digit multiplication.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Sequences and Series
