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 6
Q63
jee_main_2025_29_jan_morning
Arithmetic Progression
Consider an
A.P. of positive integers, whose \sum of the first three terms is 54 and the \sum of the first twenty terms lies between 1600 and 1800. Then its
11^textth$11^{\text{th}}$ term is:
- A. 84
- B. 122
- C. 90
- D. 108
Solution
### Related Formula
S_n = fracn2 [2a + (n-1)d]$$S_n = \frac{n}{2} [2a + (n-1)d]$$
a_n = a + (n-1)d$$a_n = a + (n-1)d$$
### Core Logic
Given S_3 = 54 implies 3a + 3d = 54 implies a + d = 18$S_3 = 54 \implies 3a + 3d = 54 \implies a + d = 18$.
Express S_20$S_{20}$ as:
S_20 = frac202[2a + 19d] = 10(2a + 19d)$$S_{20} = \frac{20}{2}[2a + 19d] = 10(2a + 19d)$$
Substitute a = 18 - d$a = 18 - d$ into the expression:
S_20 = 10[2(18 - d) + 19d] = 10(36 + 17d)$$S_{20} = 10[2(18 - d) + 19d] = 10(36 + 17d)$$
### Step 1: Formulate Inequality and Constraint Bound
Given 1600 < S_20 < 1800$1600 < S_{20} < 1800$:
1600 < 10(36 + 17d) < 1800$$1600 < 10(36 + 17d) < 1800$$
160 < 36 + 17d < 180$$160 < 36 + 17d < 180$$
124 < 17d < 144$124 < 17d < 144$
frac12417 < d < frac14417 implies 7.29 < d < 8.47$$\frac{124}{17} < d < \frac{144}{17} \implies 7.29 < d < 8.47$$
### Step 2: Isolate Integer Term parameters
Since the sequence consists of positive integers, common difference d$d$ must be an integer implies d = 8$\implies d = 8$.
Then a = 18 - 8 = 10$a = 18 - 8 = 10$.
### Step 3: Calculate the 11th Term
a_11 = a + 10d = 10 + 10(8) = 90$$a_{11} = a + 10d = 10 + 10(8) = 90$$
### Pattern Recognition
Diophantine properties (integer conditions) drastically restrict valid inequality windows. Always check parameters for strict divisibility to skip unnecessary computation.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Sequences and Series
Q24
jee_main_2024_01_february_morning
Common Terms of Two APs
Let 3, 7, 11, 15, ...., 403$3, 7, 11, 15, ...., 403$ and 2, 5, 8, 11,..., 404$2, 5, 8, 11,..., 404$ be two arithmetic progressions. Then the sum of the common terms in them is equal to
Numerical Answer. Answer: 6699 to 6699
Solution
### Related Formula
- General term of an AP: T_n = a + (n-1)d$T_n = a + (n-1)d$
- Sum of n$n$ terms of an AP: S_n = fracn2[2a + (n-1)d]$S_n = \frac{n}{2}[2a + (n-1)d]$
- The common difference of a series of common terms between two APs is given by the least common multiple of their respective common differences:
d_textcommon = textLCM(d_1, d_2)$$d_{\text{common}} = \text{LCM}(d_1, d_2)$$
### Core Logic
Let's analyze both arithmetic progressions:
- AP 1: 3, 7, 11, 15, dots, 403 implies$3, 7, 11, 15, \dots, 403 \implies$ First term a_1 = 3$a_1 = 3$, common difference d_1 = 4$d_1 = 4$.
- AP 2: 2, 5, 8, 11, dots, 404 implies$2, 5, 8, 11, \dots, 404 \implies$ First term a_2 = 2$a_2 = 2$, common difference d_2 = 3$d_2 = 3$.
By observation, the first identical value appearing in both series is 11$11$.
Therefore, the new common AP has:
- First term a = 11$a = 11$
- Common difference d = textLCM(4, 3) = 12$d = \text{LCM}(4, 3) = 12$
### Step 1: Determine the Number of Common Terms
The last term T_n$T_n$ of the common AP cannot exceed the boundary upper limits of either individual series (i.e., le 403$\le 403$):
T_n = 11 + (n-1)12 le 403$$T_n = 11 + (n-1)12 \le 403$$
12(n-1) le 392$12(n-1) \le 392$
n-1 le 32.66 implies n = 33$$n-1 \le 32.66 \implies n = 33$$
### Step 2: Calculate the Series Sum
Using the AP summation formula for 33$33$ terms:
S_33 = frac332 left[ 2(11) + (33-1)12 right]$$S_{33} = \frac{33}{2} \left[ 2(11) + (33-1)12 \right]$$
S_33 = frac332 left[ 22 + 32 times 12 right]$$S_{33} = \frac{33}{2} \left[ 22 + 32 \times 12 \right]$$
S_33 = frac332 left[ 22 + 384 right]$$S_{33} = \frac{33}{2} \left[ 22 + 384 \right]$$
S_33 = frac332 times 406 = 33 times 203 = 6699$$S_{33} = \frac{33}{2} \times 406 = 33 \times 203 = 6699$$
### Pattern Recognition
Sees: Overlapping arithmetic progression series elements.
Shortcut: Once the first matching number and the LCM are calculated, the maximum term inequality a + (n-1)d le min(L_1, L_2)$a + (n-1)d \le \min(L_1, L_2)$ directly maps out the total number of terms cleanly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Sequences and Series
Q9
jee_main_2024_29_january_evening
Arithmetic Progression
If log_mathrmea, log_mathrmeb, log_mathrmec$\log_{\mathrm{e}}a, \log_{\mathrm{e}}b, \log_{\mathrm{e}}c$ are in an A.P. and log_mathrmea - log_mathrme2b, log_mathrme2b - log_mathrme3c, log_mathrme3c - log_mathrmea$\log_{\mathrm{e}}a - \log_{\mathrm{e}}2b, \log_{\mathrm{e}}2b - \log_{\mathrm{e}}3c, \log_{\mathrm{e}}3c - \log_{\mathrm{e}}a$ are also in an A.P, then a:b:c$a:b:c$ is equal to
- A. 9 : 6 : 4
- B. 16 : 4 : 1
- C. 25 : 10 : 4
- D. 6 : 3 : 2
Solution
### Related Formula
If x, y, z$x, y, z$ are in A.P., then 2y = x + z$2y = x + z$.
### Core Logic
From the first sequence condition:
2log_e b = log_e a + log_e c implies log_e b^2 = log_e(ac) implies b^2 = ac quad dots (i)$$2\log_e b = \log_e a + \log_e c \implies \log_e b^2 = \log_e(ac) \implies b^2 = ac \quad \dots (i)$$
From the second sequence condition, the components are log_eleft(fraca2bright), log_eleft(frac2b3cright), log_eleft(frac3caright)$\log_e\left(\frac{a}{2b}\right), \log_e\left(\frac{2b}{3c}\right), \log_e\left(\frac{3c}{a}\right)$:
2log_eleft(frac2b3cright) = log_eleft(fraca2bright) + log_eleft(frac3caright)$$2\log_e\left(\frac{2b}{3c}\right) = \log_e\left(\frac{a}{2b}\right) + \log_e\left(\frac{3c}{a}\right)$$
left(frac2b3cright)^2 = fraca2b times frac3ca = frac3c2b$$\left(\frac{2b}{3c}\right)^2 = \frac{a}{2b} \times \frac{3c}{a} = \frac{3c}{2b}$$
frac4b^29c^2 = frac3c2b implies 8b^3 = 27c^3 implies fracbc = frac32 quad dots (ii)$$\frac{4b^2}{9c^2} = \frac{3c}{2b} \implies 8b^3 = 27c^3 \implies \frac{b}{c} = \frac{3}{2} \quad \dots (ii)$$
### Step 1: Finding Ratios
Substituting c = frac2b3$c = \frac{2b}{3}$ into equation (i):
b^2 = a left(frac2b3right) implies b = frac2a3 implies fracab = frac32$$b^2 = a \left(\frac{2b}{3}\right) \implies b = \frac{2a}{3} \implies \frac{a}{b} = \frac{3}{2}$$
Thus, consolidating all parts:
a : b = 9 : 6$a : b = 9 : 6$
b : c = 6 : 4$b : c = 6 : 4$
a : b : c = 9 : 6 : 4$$a : b : c = 9 : 6 : 4$$
### Pattern Recognition
Logarithmic A.P. strings immediately translate to simple geometric proportions inside the core arguments via logarithmic properties (2log x = log x^2$2\log x = \log x^2$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Sequences and Series
Q13
jee_main_2024_29_january_evening
Geometric Progression
If each term of a geometric progression a_1, a_2, a_3, ldots$a_1, a_2, a_3, \ldots$ with a_1 = frac18$a_1 = \frac{1}{8}$ and a_2 neq a_1$a_2 \neq a_1$, is the arithmetic mean of the next two terms and S_n = a_1 + a_2 + ldots + a_n$S_n = a_1 + a_2 + \ldots + a_n$, then S_20 - S_18$S_{20} - S_{18}$ is equal to
- A. 2^15$2^{15}$
- B. -2^18$-2^{18}$
- C. 2^18$2^{18}$
- D. -2^15$-2^{15}$
Solution
### Related Formula
2a_n = a_n+1 + a_n+2$$2a_n = a_{n+1} + a_{n+2}$$
### Core Logic
Let the terms of the Geometric Progression have a common ratio r$r$.
Substituting the geometric forms into the arithmetic mean relationship:
2(a r^n-1) = a r^n + a r^n+1$$2(a r^{n-1}) = a r^n + a r^{n+1}$$
Dividing out non-zero fields a r^n-1$a r^{n-1}$:
2 = r + r^2 implies r^2 + r - 2 = 0$$2 = r + r^2 \implies r^2 + r - 2 = 0$$
(r + 2)(r - 1) = 0$$(r + 2)(r - 1) = 0$$
Since a_2 neq a_1$a_2 \neq a_1$, we have r neq 1$r \neq 1$. Thus, the common ratio is r = -2$r = -2$.
### Step 1: Evaluating the Target Partial Difference
We need to evaluate:
S_20 - S_18 = T_19 + T_20$$S_{20} - S_{18} = T_{19} + T_{20}$$
T_19 + T_20 = a r^18 + a r^19 = a r^18(1 + r)$$T_{19} + T_{20} = a r^{18} + a r^{19} = a r^{18}(1 + r)$$
Substituting a = frac18$a = \frac{1}{8}$ and r = -2$r = -2$:
T_19 + T_20 = frac18 (-2)^18 (1 - 2) = frac12^3 cdot 2^18 cdot (-1) = -2^15$$T_{19} + T_{20} = \frac{1}{8} (-2)^{18} (1 - 2) = \frac{1}{2^3} \cdot 2^{18} \cdot (-1) = -2^{15}$$
### Pattern Recognition
Partial sum differences simplify into standard standalone term values (S_n - S_n-2 = T_n + T_n-1$S_n - S_{n-2} = T_n + T_{n-1}$). This eliminates the need to apply long fraction sum formats.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Sequences and Series
