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 11th$11^{\text{th}}$ term is:

Related Formula

• An Arithmetic Progression (A.P.) with common difference d$d$ sets consecutive terms as: Tₙ = T₁ + (n-1)d$T_n = T_1 + (n-1)d$
• A Geometric Progression (G.P.) ensures: T₂² = T₁ · T₃$T_2^2 = T_1 \cdot T_3$

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$3, a-1, b+1, c+9$):

Step 1: Compute the Common Difference

Using the geometric mean property for the first three terms (3, 2+d, 4+2d$3, 2+d, 4+2d$):

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:

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.

Chapter Mix

Class 11 Mathematics: Sequences and Series

Related Formula

• General term of an AP: Tₙ = a + (n-1)d$T_n = a + (n-1)d$
• Sum of n$n$ terms of an AP: Sₙ = (n)/(2)[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:

Core Logic

Let's analyze both arithmetic progressions:

• AP 1: 3, 7, 11, 15, , 403$3, 7, 11, 15, \dots, 403 \implies$ First term a₁ = 3$a_1 = 3$, common difference d₁ = 4$d_1 = 4$.
• AP 2: 2, 5, 8, 11, , 404$2, 5, 8, 11, \dots, 404 \implies$ First term a₂ = 2$a_2 = 2$, common difference d₂ = 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 = LCM(4, 3) = 12$d = \text{LCM}(4, 3) = 12$

Step 1: Determine the Number of Common Terms

The last term Tₙ$T_n$ of the common AP cannot exceed the boundary upper limits of either individual series (i.e., ≤ 403$\le 403$):

12(n-1) ≤ 392$12(n-1) \le 392$

Step 2: Calculate the Series Sum

Using the AP summation formula for 33$33$ terms:

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 ≤ (L₁, L₂)$a + (n-1)d \le \min(L_1, L_2)$ directly maps out the total number of terms cleanly.

Chapter Mix

Class 11 Mathematics: Sequences and Series

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:

From the second sequence condition, the components are ₑ((a)/(2b)), ₑ((2b)/(3c)), ₑ((3c)/(a))$\log_e\left(\frac{a}{2b}\right), \log_e\left(\frac{2b}{3c}\right), \log_e\left(\frac{3c}{a}\right)$:

Step 1: Finding Ratios

Substituting c = (2b)/(3)$c = \frac{2b}{3}$ into equation (i):

Thus, consolidating all parts: a : b = 9 : 6$a : b = 9 : 6$ b : c = 6 : 4$b : c = 6 : 4$

Pattern Recognition

Logarithmic A.P. strings immediately translate to simple geometric proportions inside the core arguments via logarithmic properties (2 x = x²$2\log x = \log x^2$).

Chapter Mix

Class 11 Mathematics: Sequences and Series

Related Formula

Core Logic

Let the terms of the Geometric Progression have a common ratio r$r$. Substituting the geometric forms into the arithmetic mean relationship:

Dividing out non-zero fields a rⁿ⁻¹$a r^{n-1}$:

Since a₂ ≠ a₁$a_2 \neq a_1$, we have r ≠ 1$r \neq 1$. Thus, the common ratio is r = -2$r = -2$.

Step 1: Evaluating the Target Partial Difference

We need to evaluate:

Substituting a = (1)/(8)$a = \frac{1}{8}$ and r = -2$r = -2$:

Pattern Recognition

Partial sum differences simplify into standard standalone term values (Sₙ - Sₙ₋₂ = Tₙ + Tₙ₋₁$S_n - S_{n-2} = T_n + T_{n-1}$). This eliminates the need to apply long fraction sum formats.

Chapter Mix

Class 11 Mathematics: Sequences and Series