Solution & Explanation
### Related Formula
- An Arithmetic Progression (A.P.) with common difference d$d$ sets consecutive terms as: T_n = T_1 + (n-1)d$T_n = T_1 + (n-1)d$
- A Geometric Progression (G.P.) ensures: T_2^2 = T_1 cdot T_3$T_2^2 = T_1 \cdot T_3$
### Core Logic
Since 3, a, b, c$3, a, b, c$ are elements of an A.P., let d$d$ denote the common difference:
- a = 3 + d$a = 3 + d$
- b = 3 + 2d$b = 3 + 2d$
- c = 3 + 3d$c = 3 + 3d$
Substituting these values into the sequence configurations of the given G.P. (3, a-1, b+1, c+9$3, a-1, b+1, c+9$):
textG.P. terms: 3, \, (3+d-1), \, (3+2d+1), \, (3+3d+9)$$\text{G.P. terms: } 3, \, (3+d-1), \, (3+2d+1), \, (3+3d+9)$$
textG.P. terms: 3, \, 2+d, \, 4+2d, \, 12+3d$$\text{G.P. terms: } 3, \, 2+d, \, 4+2d, \, 12+3d$$
### Step 1: Compute the Common Difference
Using the geometric mean property for the first three terms (3, 2+d, 4+2d$3, 2+d, 4+2d$):
(2+d)^2 = 3(4 + 2d)$$(2+d)^2 = 3(4 + 2d)$$
4 + 4d + d^2 = 12 + 6d$$4 + 4d + d^2 = 12 + 6d$$
d^2 - 2d - 8 = 0$$d^2 - 2d - 8 = 0$$
(d-4)(d+2) = 0 implies d = 4 quad textor quad d = -2$$(d-4)(d+2) = 0 \implies d = 4 \quad \text{or} \quad d = -2$$
### Step 2: Evaluate both cases for the Progressions
- **Case A: If d = 4$d = 4$**
The G.P. sequence reads: 3, 6, 12, 24$3, 6, 12, 24$ (common ratio r=2$r=2$, valid layout).
The values are: a = 7, b = 11, c = 15$a = 7, b = 11, c = 15$.
- **Case B: If d = -2$d = -2$**
The G.P. sequence reads: 3, 0, 0, 6$3, 0, 0, 6$ (contains zeros, violating standard geometric definitions).
Hence, select d = 4$d = 4$.
### Step 3: Calculate the Final Arithmetic Mean
The required arithmetic mean of a, b, c$a, b, c$ is:
textArithmetic Mean = fraca+b+c3 = frac7+11+153 = frac333 = 11$$\text{Arithmetic Mean} = \frac{a+b+c}{3} = \frac{7+11+15}{3} = \frac{33}{3} = 11$$
### Pattern Recognition
Sees: Transition parameters mapping from A.P. linear spacing into G.P. ratios.
Shortcut: Notice that the arithmetic mean of a, b, c$a, b, c$ is exactly equal to the middle value b$b$ for any linear sequence. Thus, finding b = 3 + 2(4) = 11$b = 3 + 2(4) = 11$ directly yields the final answer.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Sequences and Series
More Sequences and Series Previous-Year Questions — Page 4
Q65
jee_main_2025_04_april_evening
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
Q53
jee_main_2025_04_april_morning
Arithmetic Progression
Let A = \1, 6, 11, 16, dots\$A = \{1, 6, 11, 16, \dots\}$ and B = \9, 16, 23, 30, dots\$B = \{9, 16, 23, 30, \dots\}$ be the sets consisting of the first 2025 terms of two arithmetic progressions. Then n(A cup B)$n(A \cup B)$ is
- A. 3814
- B. 4027
- C. 3761
- D. 4003
Solution
### Related Formula
Set Principle of Inclusion-Exclusion:
n(A cup B) = n(A) + n(B) - n(A cap B)$$n(A \cup B) = n(A) + n(B) - n(A \cap B)$$
### Core Logic
Find the last terms of both progressions:
For set
A$A$:
a_1 = 1, d_1 = 5 implies T_2025 = 1 + (2025 - 1) times 5 = 10121$a_1 = 1, d_1 = 5 \implies T_{2025} = 1 + (2025 - 1) \times 5 = 10121$.
For set
B$B$:
b_1 = 9, d_2 = 7 implies T_2025 = 9 + (2025 - 1) times 7 = 14177$b_1 = 9, d_2 = 7 \implies T_{2025} = 9 + (2025 - 1) \times 7 = 14177$.
The intersection set
A cap B$A \cap B$ forms an AP with a common difference
d = textLCM(5, 7) = 35$d = \text{LCM}(5, 7) = 35$.
The first common term is
16$16$.
### Step 1: Find Common Terms Count
The general term of the common AP must satisfy:
T_n = 16 + (n - 1) times 35 le min(10121, 14177) = 10121$$T_n = 16 + (n - 1) \times 35 \le \min(10121, 14177) = 10121$$
(n - 1) times 35 le 10105 implies n - 1 le 288.71 implies n = 289$$(n - 1) \times 35 \le 10105 \implies n - 1 \le 288.71 \implies n = 289$$
### Step 2: Total Distinct Terms
Apply the inclusion-exclusion principle:
n(A cup B) = 2025 + 2025 - 289 = 3761$$n(A \cup B) = 2025 + 2025 - 289 = 3761$$
### Pattern Recognition
Common terms of two APs always generate a new AP whose common difference is the LCM of the individual common differences. Always verify the upper limit bound using the smaller of the two final values.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Sequence and Series
Q61
jee_main_2025_04_april_morning
Special Series
1 + 3 + 5^2 + 7 + 9^2 + dots$1 + 3 + 5^2 + 7 + 9^2 + \dots$ upto 40 terms is equal to
- A. 43890
- B. 41880
- C. 33980
- D. 40870
Solution
### Related Formula
Summation Identities:
sum r = fracn(n+1)2, quad sum r^2 = fracn(n+1)(2n+1)6$$\sum r = \frac{n(n+1)}{2}, \quad \sum r^2 = \frac{n(n+1)(2n+1)}{6}$$
### Core Logic
Split the 40-term series into two sub-series of 20 terms each:
Series 1 (squared terms at positions 1, 3, 5... wait, positions are odd numbers whose base squares are odd):
1^2 + 5^2 + 9^2 + dots$1^2 + 5^2 + 9^2 + \dots$ upto 20 terms. General term T_r = (4r - 3)^2$T_r = (4r - 3)^2$.
Series 2 (linear terms at positions 2, 4, 6...):
3 + 7 + 11 + dots$3 + 7 + 11 + \dots$ upto 20 terms. General term t_r = (4r - 1)$t_r = (4r - 1)$.
### Step 1: Formulate Total Sigma Expression
textSum = sum_r=1^20 left[ (4r - 3)^2 + (4r - 1) right]$$\text{Sum} = \sum_{r=1}^{20} \left[ (4r - 3)^2 + (4r - 1) \right]$$
textSum = sum_r=1^20 (16r^2 - 24r + 9 + 4r - 1) = sum_r=1^20 (16r^2 - 20r + 8)$$\text{Sum} = \sum_{r=1}^{20} (16r^2 - 24r + 9 + 4r - 1) = \sum_{r=1}^{20} (16r^2 - 20r + 8)$$
textSum = 16sum_r=1^20 r^2 - 20sum_r=1^20 r + 8sum_r=1^20 1$$\text{Sum} = 16\sum_{r=1}^{20} r^2 - 20\sum_{r=1}^{20} r + 8\sum_{r=1}^{20} 1$$
### Step 2: Arithmetic Evaluation
sum_r=1^20 r^2 = frac20 times 21 times 416 = 2870$$\sum_{r=1}^{20} r^2 = \frac{20 \times 21 \times 41}{6} = 2870$$
sum_r=1^20 r = frac20 times 212 = 210$$\sum_{r=1}^{20} r = \frac{20 \times 21}{2} = 210$$
textSum = 16(2870) - 20(210) + 8(20) = 45920 - 4200 + 160 = 41880$$\text{Sum} = 16(2870) - 20(210) + 8(20) = 45920 - 4200 + 160 = 41880$$
### Pattern Recognition
When dealing with interlaced series, pairing terms adjacent to each other simplifies the degree of general expressions into manageable standard summation polynomials.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Sequence and Series
Q60
jee_main_2025_07_april_evening
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
Q70
jee_main_2025_07_april_evening
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
