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
null
### Chapter Mix
Class 11 Mathematics: Sequences and Series
More Sequences and Series Previous-Year Questions — Page 3
Q66
2025
Geometric Progression
Let a_1, a_2, a_3, ldots$a_1, a_2, a_3, \ldots$ be a G.P. of increasing positive numbers[cite: 655]. If a_3a_5 = 729$a_3a_5 = 729$ and a_2 + a_4 = frac1114$a_2 + a_4 = \frac{111}{4}$ [cite: 656], then 24(a_1 + a_2 + a_3)$24(a_1 + a_2 + a_3)$ is equal to[cite: 657]:
- A. 131
- B. 130
- C. 129
- D. 128
Solution
### Related Formula
For a geometric sequence configuration with first term a$a$ and common ratio r$r$:
a_n = a cdot r^n-1$a_n = a \cdot r^{n-1}$
### Core Logic
Convert information markers using parameter notations [cite: 1372, 1373, 1376]:
a_3 a_5 = (ar^2)(ar^4) = a^2 r^6 = 729 implies ar^3 = 27$a_3 a_5 = (ar^2)(ar^4) = a^2 r^6 = 729 \implies ar^3 = 27$ [cite: 1373, 1374]
From second expression block data [cite: 1376]:
a_2 + a_4 = ar + ar^3 = frac1114$a_2 + a_4 = ar + ar^3 = \frac{111}{4}$ [cite: 1376]
Substitute ar^3 = 27$ar^3 = 27$ directly into the linear equation block [cite: 1376]:
ar + 27 = frac1114 implies ar = frac1114 - 27 = frac34$ar + 27 = \frac{111}{4} \implies ar = \frac{111}{4} - 27 = \frac{3}{4}$ [cite: 1376]
### Step 1: Finding parameters a and r
Divide the calculated components to evaluate the ratio [cite: 1387]:
fracar^3ar = frac273/4 implies r^2 = 36 implies r = 6$\frac{ar^3}{ar} = \frac{27}{3/4} \implies r^2 = 36 \implies r = 6$ [cite: 1387]
(Choose +6$+6$ because terms must stay strictly positive [cite: 655]).
Find value for first base variable a$a$ [cite: 1389]:
a(6) = frac34 implies a = frac18$a(6) = \frac{3}{4} \implies a = \frac{1}{8}$ [cite: 1389]
### Step 2: Sum configuration resolving
Now compute targeted expansion expression value [cite: 1390]:
24(a_1 + a_2 + a_3) = 24(a + ar + ar^2) = 24a(1 + r + r^2)$24(a_1 + a_2 + a_3) = 24(a + ar + ar^2) = 24a(1 + r + r^2)$ [cite: 1390]
= 24 cdot left(frac18right) cdot (1 + 6 + 36) = 3 cdot 43 = 129$= 24 \cdot \left(\frac{1}{8}\right) \cdot (1 + 6 + 36) = 3 \cdot 43 = 129$ [cite: 1390, 1391]
### Pattern Recognition
Product entries like a_3 a_5 = a_4^2$a_3 a_5 = a_4^2$ help identify the central term index value quickly in symmetric geometric progressions.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Sequences and Series
Q62
2025
Telescoping Series
If the sum of the first 20 terms of the series frac 4 . 14 + 3 . 1 ^ 2 + 1 ^ 4 + frac 4 . 24 + 3 . 2 ^ 2 + 2 ^ 4 + frac 4 . 34 + 3 . 3 ^ 2 + 3 ^ 4 + frac 4 . 44 + 3 . 4 ^ 2 + 4 ^ 4 + dots$\frac {4 . 1}{4 + 3 . 1 ^ {2} + 1 ^ {4}} + \frac {4 . 2}{4 + 3 . 2 ^ {2} + 2 ^ {4}} + \frac {4 . 3}{4 + 3 . 3 ^ {2} + 3 ^ {4}} + \frac {4 . 4}{4 + 3 . 4 ^ {2} + 4 ^ {4}} + \dots$ is fracmathrmmmathrmn$\frac{\mathrm{m}}{\mathrm{n}}$, where m and n are coprime, then mathrmm + mathrmn$\mathrm{m} + \mathrm{n}$ is equal to:
- A. 423$423$
- B. 420$420$
- C. 421$421$
- D. 422$422$
Solution
### Core Logic
The general term T_r$T_r$ of the series can be written as:
T_r = frac4rr^4 + 3r^2 + 4$T_r = \frac{4r}{r^4 + 3r^2 + 4}$
Let's factorize the denominator by completing the square metric:
r^4 + 3r^2 + 4 = (r^4 + 4r^2 + 4) - r^2 = (r^2 + 2)^2 - r^2$r^4 + 3r^2 + 4 = (r^4 + 4r^2 + 4) - r^2 = (r^2 + 2)^2 - r^2$
Using the difference of squares identity A^2 - B^2 = (A-B)(A+B)$A^2 - B^2 = (A-B)(A+B)$:
r^4 + 3r^2 + 4 = (r^2 - r + 2)(r^2 + r + 2)$r^4 + 3r^2 + 4 = (r^2 - r + 2)(r^2 + r + 2)$
### Step 1: Partial Fraction Decomposition
Express T_r$T_r$ using partial fractions split:
T_r = frac4r(r^2 - r + 2)(r^2 + r + 2) = 2 left[ frac1r^2 - r + 2 - frac1r^2 + r + 2 right]$T_r = \frac{4r}{(r^2 - r + 2)(r^2 + r + 2)} = 2 \left[ \frac{1}{r^2 - r + 2} - \frac{1}{r^2 + r + 2} \right]$
Notice that if we define V(r) = r^2 - r + 2$V(r) = r^2 - r + 2$, then V(r+1) = (r+1)^2 - (r+1) + 2 = r^2 + 2r + 1 - r - 1 + 2 = r^2 + r + 2$V(r+1) = (r+1)^2 - (r+1) + 2 = r^2 + 2r + 1 - r - 1 + 2 = r^2 + r + 2$.
Thus, T_r = 2big[V(r) - V(r+1)big]$T_r = 2\big[V(r) - V(r+1)\big]$, which sets up a clear telescoping sum formulation.
### Step 2: Evaluating the Sum of 20 Terms
Summing from r = 1$r = 1$ to 20$20$:
S_20 = sum_r=1^20 T_r = 2 sum_r=1^20 left[ frac1r^2 - r + 2 - frac1r^2 + r + 2 right]$S_{20} = \sum_{r=1}^{20} T_r = 2 \sum_{r=1}^{20} \left[ \frac{1}{r^2 - r + 2} - \frac{1}{r^2 + r + 2} \right]$
= 2 left[ left(frac12 - frac14right) + left(frac14 - frac18right) + dots + left(frac120^2 - 20 + 2 - frac120^2 + 20 + 2right) right]$= 2 \left[ \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{8}\right) + \dots + \left(\frac{1}{20^2 - 20 + 2} - \frac{1}{20^2 + 20 + 2}\right) \right]$
All sequential middle terms cancel completely, leaving only first and final values:
S_20 = 2 left[ frac12 - frac1422 right] = 1 - frac1211 = frac210211$S_{20} = 2 \left[ \frac{1}{2} - \frac{1}{422} \right] = 1 - \frac{1}{211} = \frac{210}{211}$
Since 210$210$ and 211$211$ are coprime, m = 210$m = 210$ and n = 211$n = 211$.
### Step 3: Calculating m + n
Combining both values:
m + n = 210 + 211 = 421$m + n = 210 + 211 = 421$
### Pattern Recognition
The polynomial factorization r^4 + a^2r^2 + b^4$r^4 + a^2r^2 + b^4$ is a frequent pattern in series problems. Always complete the square to break it into a product of quadratic expressions, which naturally yields a telescoping sequence.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Sequences and Series
Q65
2025
Arithmetic Progression
Consider two sets A and B, each containing three numbers in A.P. Let the sum and the product of the elements of A be 36 and p respectively and the sum and the product of the elements of B be 36 and q respectively. Let d and D be the common differences of AP's in A and B respectively such that mathrmD = mathrmd + 3, mathrm~d > 0$\mathrm{D} = \mathrm{d} + 3, \mathrm{~d} > 0$. If fracmathrmp + mathrmqmathrmp - mathrmq = frac195$\frac{\mathrm{p} + \mathrm{q}}{\mathrm{p} - \mathrm{q}} = \frac{19}{5}$, then mathrmp - mathrmq$\mathrm{p} - \mathrm{q}$ is equal to
- A. 600$600$
- B. 450$450$
- C. 630$630$
- D. 540$540$
Solution
### Core Logic
Let the 3 elements of set A$A$ in A.P. be a-d, a, a+d$a-d, a, a+d$.
Their sum is 3a = 36 implies a = 12$3a = 36 \implies a = 12$.
Their product is p = a(a^2 - d^2) = 12(144 - d^2)$p = a(a^2 - d^2) = 12(144 - d^2)$.
Similarly, let the 3 elements of set B$B$ be b-D, b, b+D$b-D, b, b+D$.
Their sum is 3b = 36 implies b = 12$3b = 36 \implies b = 12$.
Their product is q = b(b^2 - D^2) = 12(144 - D^2)$q = b(b^2 - D^2) = 12(144 - D^2)$.
### Step 1: Using the Ratio Condition
We are given the relation:
fracp + qp - q = frac195$\frac{p + q}{p - q} = \frac{19}{5}$
Using componendo and dividendo:
fracpq = frac19 + 519 - 5 = frac2414 = frac127$\frac{p}{q} = \frac{19 + 5}{19 - 5} = \frac{24}{14} = \frac{12}{7}$
Substitute the expression blocks for p$p$ and q$q$:
frac12(144 - d^2)12(144 - D^2) = frac127 implies frac144 - d^2144 - D^2 = frac127$\frac{12(144 - d^2)}{12(144 - D^2)} = \frac{12}{7} \implies \frac{144 - d^2}{144 - D^2} = \frac{12}{7}$
7(144 - d^2) = 12(144 - D^2)$7(144 - d^2) = 12(144 - D^2)$
### Step 2: Substituting D in terms of d
We are given D = d + 3$D = d + 3$:
7(144 - d^2) = 12big(144 - (d + 3)^2big)$7(144 - d^2) = 12\big(144 - (d + 3)^2\big)$
1008 - 7d^2 = 12big(144 - (d^2 + 6d + 9)big)$1008 - 7d^2 = 12\big(144 - (d^2 + 6d + 9)\big)$
1008 - 7d^2 = 12big(135 - d^2 - 6dbig) = 1620 - 12d^2 - 72d$1008 - 7d^2 = 12\big(135 - d^2 - 6d\big) = 1620 - 12d^2 - 72d$
5d^2 + 72d - 612 = 0$5d^2 + 72d - 612 = 0$
Solving this quadratic equation:
(d - 6)(5d + 102) = 0$(d - 6)(5d + 102) = 0$
Since d > 0$d > 0$, we choose d = 6$d = 6$. This implies D = 6 + 3 = 9$D = 6 + 3 = 9$.
### Step 3: Finding p - q
Now calculate the targeted metric:
p - q = 12(144 - d^2) - 12(144 - D^2) = 12(D^2 - d^2)$p - q = 12(144 - d^2) - 12(144 - D^2) = 12(D^2 - d^2)$
p - q = 12(9^2 - 6^2) = 12(81 - 36) = 12(45) = 540$p - q = 12(9^2 - 6^2) = 12(81 - 36) = 12(45) = 540$
### Pattern Recognition
For 3-element symmetric AP sequences, choosing terms as x-d, x, x+d$x-d, x, x+d$ ensures the sum isolates the middle term instantly (3x = S$3x = S$). This drastically drops algebraic variables from the start.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Sequences and Series
Q60
2025
Arithmetic Progression
Let a_n$a_n$ be the n^textth$n^{\text{th}}$ term of an A. P. If S_mathrmn = a_1 + a_2 + a_3 + dots + a_mathrmn = 700, a_6 = 7$S_{\mathrm{n}} = a_{1} + a_{2} + a_{3} + \dots + a_{\mathrm{n}} = 700, a_{6} = 7$ and S_7 = 7$S_7 = 7$, then a_n$a_n$ is equal to:
- A. 56$56$
- B. 65$65$
- C. 64$64$
- D. 70$70$
Solution
### Related Formula
Sum of first n$n$ terms of an AP is given by:
S_n = fracn2[2a + (n-1)d]$S_n = \frac{n}{2}[2a + (n-1)d]$
### Core Logic
Given specifications:
1) a_6 = 7 implies a + 5d = 7 quad dots text(ii)$a_6 = 7 \implies a + 5d = 7 \quad \dots \text{(ii)}$
2) S_7 = 7 implies frac72(2a + 6d) = 7 implies a + 3d = 1 quad dots text(iii)$S_7 = 7 \implies \frac{7}{2}(2a + 6d) = 7 \implies a + 3d = 1 \quad \dots \text{(iii)}$
Subtracting (iii) from (ii):
2d = 6 implies d = 3$2d = 6 \implies d = 3$
Substituting d=3$d=3$ into (iii):
a + 3(3) = 1 implies a = -8$a + 3(3) = 1 \implies a = -8$
### Step 1: Find n from Sn = 700
Substitute a = -8$a = -8$ and d = 3$d = 3$ into the equation for S_n = 700$S_n = 700$:
700 = fracn2[2(-8) + (n-1)3]$700 = \frac{n}{2}[2(-8) + (n-1)3]$
1400 = n[-16 + 3n - 3]$1400 = n[-16 + 3n - 3]$
3n^2 - 19n - 1400 = 0$3n^2 - 19n - 1400 = 0$
Factoring the quadratic equation:
(3n + 56)(n - 25) = 0$(3n + 56)(n - 25) = 0$
Since n$n$ must be a positive integer, n = 25$n = 25$.
### Step 2: Determine standard term value
We need to find a_25$a_{25}$ corresponding to index n=25$n=25$:
a_25 = a + 24d$a_{25} = a + 24d$
a_25 = -8 + 24(3) = -8 + 72 = 64$a_{25} = -8 + 24(3) = -8 + 72 = 64$
### Pattern Recognition
When S_n$S_n$ and specific terms are given, prioritize finding the first term a$a$ and common difference d$d$ through simple elimination headers before targeting the value of n$n$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Sequences and Series
