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 8
Q25
jee_main_2024_30_january_evening
Arithmetic Progression
Let S_n$S_n$ be the sum to n$n$-terms of an arithmetic progression 3, 7, 11, dots$3, 7, 11, \dots$
If 40 lt left(frac6n(n + 1)sum_k=1^nS_kright) lt 42$40 \lt \left(\frac{6}{n(n + 1)}\sum_{k=1}^{n}S_k\right) \lt 42$ , then n$n$ equals
Numerical Answer. Answer: 9 to 9
Solution
### Related Formula
textSum of AP: S_k = frack2 [2a + (k - 1)d]$$\text{Sum of AP: } S_k = \frac{k}{2} [2a + (k - 1)d]$$
sum_k=1^n k^2 = fracn(n+1)(2n+1)6$$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$$
sum_k=1^n k = fracn(n+1)2$$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$$
### Core Logic
For the arithmetic progression 3, 7, 11, dots$3, 7, 11, \dots$
First term a = 3$a = 3$, Common difference d = 4$d = 4$.
The sum of the first k$k$ terms is:
S_k = frack2 (2(3) + (k - 1)4) = frack2 (6 + 4k - 4) = frack2 (4k + 2) = 2k^2 + k$$S_k = \frac{k}{2} (2(3) + (k - 1)4) = \frac{k}{2} (6 + 4k - 4) = \frac{k}{2} (4k + 2) = 2k^2 + k$$
### Step 1: Finding the Sum of Sums
Now compute the sum sum_k=1^n S_k$\sum_{k=1}^{n} S_k$:
sum_k=1^n S_k = sum_k=1^n (2k^2 + k) = 2sum_k=1^n k^2 + sum_k=1^n k$$\sum_{k=1}^{n} S_k = \sum_{k=1}^{n} (2k^2 + k) = 2\sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k$$
= 2 left( fracn(n+1)(2n+1)6 right) + fracn(n+1)2$$= 2 \left( \frac{n(n+1)(2n+1)}{6} \right) + \frac{n(n+1)}{2}$$
= n(n+1) left[ frac2(2n+1)6 + frac12 right] = n(n+1) left[ frac2n+13 + frac12 right]$$= n(n+1) \left[ \frac{2(2n+1)}{6} + \frac{1}{2} \right] = n(n+1) \left[ \frac{2n+1}{3} + \frac{1}{2} \right]$$
= n(n+1) left[ frac4n + 2 + 36 right] = fracn(n+1)(4n + 5)6$$= n(n+1) \left[ \frac{4n + 2 + 3}{6} \right] = \frac{n(n+1)(4n + 5)}{6}$$
### Step 2: Resolving the Inequality
Substitute this sum into the given expression:
frac6n(n+1) sum_k=1^n S_k = frac6n(n+1) cdot fracn(n+1)(4n+5)6 = 4n + 5$$\frac{6}{n(n+1)} \sum_{k=1}^{n} S_k = \frac{6}{n(n+1)} \cdot \frac{n(n+1)(4n+5)}{6} = 4n + 5$$
We are given the bounds:
40 lt 4n + 5 lt 42$$40 \lt 4n + 5 \lt 42$$
35 lt 4n lt 37$$35 \lt 4n \lt 37$$
8.75 lt n lt 9.25$$8.75 \lt n \lt 9.25$$
Since n$n$ must be an integer (representing the number of terms), the only valid integer is n = 9$n = 9$.
### Pattern Recognition
Evaluating a 'sum of sums' for an AP effectively requires applying the Sigma k^2$\Sigma k^2$ and Sigma k$\Sigma k$ standard formulas to the generic S_n$S_n$ quadratic.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Sequences and Series
Q2
jee_main_2024_30_jan_morning
Sum of n terms of AP
Let S_n$S_n$ denote the sum of first n$n$ terms an arithmetic progression. If S_20 = 790$S_{20} = 790$ and S_10 = 145$S_{10} = 145$, then S_15 - S_5$S_{15} - S_5$ is:
- A. 395$395$
- B. 390$390$
- C. 405$405$
- D. 410$410$
Solution
### Related Formula
S_n = fracn2[2a + (n-1)d]$$S_n = \frac{n}{2}[2a + (n-1)d]$$
### Core Logic
Using the sum formula for an AP:
S_20 = frac202[2a + 19d] = 790$$S_{20} = \frac{20}{2}[2a + 19d] = 790$$
10[2a + 19d] = 790$$10[2a + 19d] = 790$$
2a + 19d = 79 quad dots (1)$$2a + 19d = 79 \quad \dots (1)$$
S_10 = frac102[2a + 9d] = 145$$S_{10} = \frac{10}{2}[2a + 9d] = 145$$
5[2a + 9d] = 145$$5[2a + 9d] = 145$$
2a + 9d = 29 quad dots (2)$$2a + 9d = 29 \quad \dots (2)$$
### Step 1: Solving for a and d
Subtracting (2) from (1):
10d = 50 Rightarrow d = 5$$10d = 50 \Rightarrow d = 5$$
Substituting d=5$d=5$ into (2):
2a + 9(5) = 29 Rightarrow 2a = 29 - 45 = -16$$2a + 9(5) = 29 \Rightarrow 2a = 29 - 45 = -16$$
a = -8$a = -8$
### Step 2: Evaluating the required expression
We need to find S_15 - S_5$S_{15} - S_{5}$:
S_15 - S_5 = frac152[2a + 14d] - frac52[2a + 4d]$$S_{15} - S_5 = \frac{15}{2}[2a + 14d] - \frac{5}{2}[2a + 4d]$$
Substituting 2a = -16$2a = -16$ and d = 5$d = 5$:
= frac152[-16 + 70] - frac52[-16 + 20]$$= \frac{15}{2}[-16 + 70] - \frac{5}{2}[-16 + 20]$$
= frac152[54] - frac52[4]$$= \frac{15}{2}[54] - \frac{5}{2}[4]$$
= 15 times 27 - 5 times 2$$= 15 \times 27 - 5 \times 2$$
= 405 - 10 = 395$$= 405 - 10 = 395$$
### Pattern Recognition
When two sums of an AP are given, immediately set up the linear equations in terms of a$a$ and d$d$. Solve for them, and substitute directly into the target expression.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Sequences and Series
Q30
jee_main_2024_30_jan_morning
Special Series
Let alpha = 1^2 + 4^2 + 8^2 + 13^2 + 19^2 + 26^2 + dots$\alpha = 1^2 + 4^2 + 8^2 + 13^2 + 19^2 + 26^2 + \dots$ upto 10 terms and beta = sum_n=1^10 n^4$\beta = \sum_{n=1}^{10} n^4$. If 4alpha - beta = 55k + 40$4\alpha - \beta = 55k + 40$, then k is equal to
Numerical Answer. Answer: 353 to 353
Solution
### Related Formula
textMethod of differences for a sequence: V_n - V_n-1 = T_n$$\text{Method of differences for a sequence: } V_n - V_{n-1} = T_n$$
### Core Logic
The base terms inside the squares form a sequence: 1, 4, 8, 13, 19, 26 dots$1, 4, 8, 13, 19, 26 \dots$
The differences between consecutive terms are: 3, 4, 5, 6, 7 dots$3, 4, 5, 6, 7 \dots$
Since the first differences are an Arithmetic Progression, the general term of the inner sequence is a quadratic in n$n$: T_n = an^2 + bn + c$T_n = an^2 + bn + c$.
Using n=1$n=1$: 1 = a + b + c$1 = a + b + c$
Using n=2$n=2$: 4 = 4a + 2b + c$4 = 4a + 2b + c$
Using n=3$n=3$: 8 = 9a + 3b + c$8 = 9a + 3b + c$
Solving this system:
(4a + 2b + c) - (a + b + c) = 3 Rightarrow 3a + b = 3$(4a + 2b + c) - (a + b + c) = 3 \Rightarrow 3a + b = 3$
(9a + 3b + c) - (4a + 2b + c) = 4 Rightarrow 5a + b = 4$(9a + 3b + c) - (4a + 2b + c) = 4 \Rightarrow 5a + b = 4$
Subtracting these gives: 2a = 1 Rightarrow a = 1/2$2a = 1 \Rightarrow a = 1/2$.
Then 3(1/2) + b = 3 Rightarrow b = 3/2$3(1/2) + b = 3 \Rightarrow b = 3/2$.
Finally 1/2 + 3/2 + c = 1 Rightarrow c = -1$1/2 + 3/2 + c = 1 \Rightarrow c = -1$.
Inner sequence T_n = frac12n^2 + frac32n - 1$T_n = \frac{1}{2}n^2 + \frac{3}{2}n - 1$.
### Step 1: Calculating alpha structure
The series is alpha = sum_n=1^10 (T_n)^2$\alpha = \sum_{n=1}^{10} (T_n)^2$.
4alpha = sum_n=1^10 4left(fracn^2 + 3n - 22right)^2 = sum_n=1^10 (n^2 + 3n - 2)^2$$4\alpha = \sum_{n=1}^{10} 4\left(\frac{n^2 + 3n - 2}{2}\right)^2 = \sum_{n=1}^{10} (n^2 + 3n - 2)^2$$
Expand the squared trinomial:
(n^2 + 3n - 2)^2 = n^4 + 9n^2 + 4 + 6n^3 - 4n^2 - 12n = n^4 + 6n^3 + 5n^2 - 12n + 4$$(n^2 + 3n - 2)^2 = n^4 + 9n^2 + 4 + 6n^3 - 4n^2 - 12n = n^4 + 6n^3 + 5n^2 - 12n + 4$$
### Step 2: Applying given target relation
We are given beta = sum_n=1^10 n^4$\beta = \sum_{n=1}^{10} n^4$.
So, 4alpha - beta = sum_n=1^10 (n^4 + 6n^3 + 5n^2 - 12n + 4) - sum_n=1^10 n^4$4\alpha - \beta = \sum_{n=1}^{10} (n^4 + 6n^3 + 5n^2 - 12n + 4) - \sum_{n=1}^{10} n^4$
4alpha - beta = sum_n=1^10 (6n^3 + 5n^2 - 12n + 4)$$4\alpha - \beta = \sum_{n=1}^{10} (6n^3 + 5n^2 - 12n + 4)$$
### Step 3: Calculating summation limits
Evaluate each standard summation up to n=10$n=10$:
sum n^3 = (10 times 11 / 2)^2 = 55^2 = 3025$\sum n^3 = (10 \times 11 / 2)^2 = 55^2 = 3025$
sum n^2 = (10 times 11 times 21) / 6 = 385$\sum n^2 = (10 \times 11 \times 21) / 6 = 385$
sum n = (10 times 11) / 2 = 55$\sum n = (10 \times 11) / 2 = 55$
sum 4 = 40$\sum 4 = 40$
4alpha - beta = 6(3025) + 5(385) - 12(55) + 40$$4\alpha - \beta = 6(3025) + 5(385) - 12(55) + 40$$
= 18150 + 1925 - 660 + 40 = 19455$$= 18150 + 1925 - 660 + 40 = 19455$$
We are given 4alpha - beta = 55k + 40$4\alpha - \beta = 55k + 40$.
19455 = 55k + 40$$19455 = 55k + 40$$
19415 = 55k$19415 = 55k$
k = frac1941555 = 353$$k = \frac{19415}{55} = 353$$
### Pattern Recognition
Recognizing arithmetic progressions in the first-order differences immediately specifies a quadratic general term An^2+Bn+C$An^2+Bn+C$. Expanding and cancelling highest-order summation terms drastically simplifies standard sums.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Sequences and Series
Q9
jee_main_2024_31_jan_evening
Arithmetic and Geometric Progression
Let 2^mathrmnd$2^{\mathrm{nd}}$, 8^mathrmth$8^{\mathrm{th}}$ and 44^mathrmth$44^{\mathrm{th}}$, terms of a non-constant A.P. be respectively the 1^mathrmst$1^{\mathrm{st}}$, 2^mathrmnd$2^{\mathrm{nd}}$ and 3^mathrmrd$3^{\mathrm{rd}}$ terms of G.P. If the first term of A.P. is 1 then the sum of first 20 terms is equal to
- A. 980$980$
- B. 960$960$
- C. 990$990$
- D. 970$970$
Solution
### Related Formula
S_n = fracn2[2a + (n-1)d]$$S_n = \frac{n}{2}[2a + (n-1)d]$$
textIf p, q, r text are in G.P. then q^2 = pr$$\text{If } p, q, r \text{ are in G.P. then } q^2 = pr$$
### Core Logic
Let the A.P. be a, a+d, a+2d, dots$a, a+d, a+2d, \dots$
Given a=1$a=1$, the 2^textnd, 8^textth$2^{\text{nd}}, 8^{\text{th}}$, and 44^textth$44^{\text{th}}$ terms are:
T_2 = 1 + d$T_2 = 1 + d$
T_8 = 1 + 7d$T_8 = 1 + 7d$
T_44 = 1 + 43d$$T_{44} = 1 + 43d$$
These terms are in G.P., so:
(1+7d)^2 = (1+d)(1+43d)$$(1+7d)^2 = (1+d)(1+43d)$$
1 + 14d + 49d^2 = 1 + 44d + 43d^2$$1 + 14d + 49d^2 = 1 + 44d + 43d^2$$
6d^2 - 30d = 0 implies 6d(d - 5) = 0$$6d^2 - 30d = 0 \implies 6d(d - 5) = 0$$
Since it is a non-constant A.P., d neq 0$d \neq 0$, so d = 5$d = 5$.
Sum of first 20 terms:
S_20 = frac202[2(1) + (20-1)5]$$S_{20} = \frac{20}{2}[2(1) + (20-1)5]$$
S_20 = 10[2 + 19(5)] = 10[2 + 95] = 970$$S_{20} = 10[2 + 19(5)] = 10[2 + 95] = 970$$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Sequences and Series
Q13
jee_main_2024_31_jan_morning
Method of Differences
The sum of the series frac11 - 3 cdot 1^2 + 1^4 + frac21 - 3 cdot 2^2 + 2^4 + frac31 - 3 cdot 3^2 + 3^4 + dots$\frac{1}{1 - 3 \cdot 1^2 + 1^4} + \frac{2}{1 - 3 \cdot 2^2 + 2^4} + \frac{3}{1 - 3 \cdot 3^2 + 3^4} + \dots$ up to 10 terms is
- A. frac45109$$\frac{45}{109}$$
- B. -frac45109$$-\frac{45}{109}$$
- C. frac55109$$\frac{55}{109}$$
- D. -frac55109$$-\frac{55}{109}$$
Solution
### Core Logic
General term
T_r = fracrr^4 - 3r^2 + 1$T_r = \frac{r}{r^4 - 3r^2 + 1}$.
Factorize the denominator:
r^4 - 3r^2 + 1 = (r^4 - 2r^2 + 1) - r^2 = (r^2 - 1)^2 - r^2$$r^4 - 3r^2 + 1 = (r^4 - 2r^2 + 1) - r^2 = (r^2 - 1)^2 - r^2$$
= (r^2 - r - 1)(r^2 + r - 1)$$= (r^2 - r - 1)(r^2 + r - 1)$$
### Step 1: Partial Fractions
T_r = fracr(r^2 - r - 1)(r^2 + r - 1)$$T_r = \frac{r}{(r^2 - r - 1)(r^2 + r - 1)}$$
Notice that
(r^2 + r - 1) - (r^2 - r - 1) = 2r$(r^2 + r - 1) - (r^2 - r - 1) = 2r$.
T_r = frac12 left[ frac2r(r^2 - r - 1)(r^2 + r - 1) right] = frac12 left[ frac1r^2 - r - 1 - frac1r^2 + r - 1 right]$$T_r = \frac{1}{2} \left[ \frac{2r}{(r^2 - r - 1)(r^2 + r - 1)} \right] = \frac{1}{2} \left[ \frac{1}{r^2 - r - 1} - \frac{1}{r^2 + r - 1} \right]$$
### Step 2: Telescoping Sum
Sum
S = sum_r=1^10 T_r$S = \sum_{r=1}^{10} T_r$.
The terms will telescope because the second term for
r$r$ is identical to the first term for
r+1$r+1$. (Let
v_r = r^2 - r - 1$v_r = r^2 - r - 1$, then
v_r+1 = (r+1)^2 - (r+1) - 1 = r^2 + 2r + 1 - r - 1 - 1 = r^2 + r - 1$v_{r+1} = (r+1)^2 - (r+1) - 1 = r^2 + 2r + 1 - r - 1 - 1 = r^2 + r - 1$).
S = frac12 left[ frac11^2 - 1 - 1 - frac110^2 + 10 - 1 right]$$S = \frac{1}{2} \left[ \frac{1}{1^2 - 1 - 1} - \frac{1}{10^2 + 10 - 1} \right]$$
S = frac12 left[ frac1-1 - frac1109 right] = frac12 left[ -1 - frac1109 right] = -frac55109$$S = \frac{1}{2} \left[ \frac{1}{-1} - \frac{1}{109} \right] = \frac{1}{2} \left[ -1 - \frac{1}{109} \right] = -\frac{55}{109}$$
### Pattern Recognition
Expressions like
r^4 + kr^2 + 1$r^4 + kr^2 + 1$ can be factorized by completing the square to create a difference of two squares. This setup invariably leads to a
telescoping series.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Sequences and Series
