Let Tᵣ$T_r$ be the rth$r^{\text{th}}$ term of an A.P. If for some m, Tm = (1)/(25), T₂₅ = (1)/(20)$T_m = \frac{1}{25}, T_{25} = \frac{1}{20}$ and 20Σr = 1²⁵ Tᵣ = 13$20\sum_{r = 1}^{25} T_r = 13$ then 5mΣr = m2m Tᵣ$5m\sum_{r = m}^{2m} T_r$ is equal to:

Related Formula

Core Logic

First Progression (S₁$S_1$): 4, 9, 14, 19,$4, 9, 14, 19, \dots$ Common difference d₁ = 5$d_1 = 5$. Last term (T₂₅$T_{25}$) = 4 + (25-1)5 = 4 + 120 = 124$= 4 + (25-1)5 = 4 + 120 = 124$.

Second Progression (S₂$S_2$): 3, 6, 9, 12,$3, 6, 9, 12, \dots$ Common difference d₂ = 3$d_2 = 3$. Last term (T₃₇$T_{37}$) = 3 + (37-1)3 = 3 + 108 = 111$= 3 + (37-1)3 = 3 + 108 = 111$.

Step 1: Forming the Common AP

By inspecting the sequences, the first common term (acommon$a_{\text{common}}$) is 9$9$. The common difference of the new series is the LCM of the original differences:

Thus, the common terms form a new AP: 9, 24, 39, 54,$9, 24, 39, 54, \dots$

Step 2: Bounding the Sequence

The last term of the common AP must be less than or equal to the smallest maximum limit of the two series. Here, (124, 111) = 111$\min(124, 111) = 111$. So, the n$n$-th term of the common sequence is bounded by 111:

15(n-1) ≤ 102$15(n-1) \le 102$

n ≤ 7.8$n \le 7.8$ Since n$n$ must be an integer, n = 7$n = 7$.

Pattern Recognition

The common terms of two APs always form a new AP. Its common difference is the LCM of the original differences. Find the first common term manually, then cap the n$n$-th term inequality with the smallest end-boundary of the original sets.

Chapter Mix

Class 11 Maths: Sequences and Series

Related Formula

(Sum of an infinite Arithmetico-Geometric Progression, where a$a$ is the first AP term, d$d$ is common difference, and r$r$ is geometric ratio).

Core Logic

The series given is an AGP. However, let's look at it explicitly. Let S = 8$S = 8$.

Multiply the entire equation by the geometric ratio (1/4$1/4$):

Step 1: Shift and Subtract

Subtract the shifted series from the original series:

Step 2: Summing the pure Infinite GP

The term in parentheses is an infinite geometric series with a=1$a=1$ and r=1/4$r=1/4$. Sum = (1)/(1 - 1/4) = (1)/(3/4) = (4)/(3)$= \frac{1}{1 - 1/4} = \frac{1}{3/4} = \frac{4}{3}$

Step 3: Final Output Evaluation

Substitute this sum back:

Pattern Recognition

The shift-and-subtract technique natively nullifies the arithmetic growth leaving behind a uniform geometric progression. Using the AGP direct formula S = a/(1-r) + dr/(1-r)²$S = a/(1-r) + dr/(1-r)^2$ works perfectly here as well.

Chapter Mix

Class 11 Maths: Sequences and Series

Related Formula

where Sₙ$S_n$ is the sum of n$n$ terms, a$a$ is the first term, and r$r$ is the common ratio.

Core Logic

Let the terms of the G.P. be a, ar, ar², ar³, , ar⁶³$a, ar, ar^2, ar^3, \dots, ar^{63}$.

The sum of all 64 terms is given by:

The odd terms are a, ar², ar⁴, , ar⁶²$a, ar^2, ar^4, \dots, ar^{62}$. This forms another G.P. with 32 terms and a common ratio of r²$r^2$. The sum of the odd terms is:

Step 1: Equate and Solve for r

We are given that Sall = 7 · Sodd$S_{\text{all}} = 7 \cdot S_{\text{odd}}$. Substituting our formulas:

Assuming a ≠ 0$a \neq 0$ and r ≠ 1$r \neq 1$, we can cancel the common terms a(1 - r⁶⁴)$\dots a(1 - r^{64}) \dots$ from both sides:

Since 1 - r² = (1 - r)(1 + r)$1 - r^2 = (1 - r)(1 + r)$, we have:

1 + r = 7$1 + r = 7$

r = 6$r = 6$

Pattern Recognition

Shortcut: In any G.P. with an even number of terms, the ratio of the total sum to the sum of the odd-positioned terms is exactly 1 + r$1 + r$. Thus, 1 + r = 7 ⇒ r = 6$1 + r = 7 \Rightarrow r = 6$ immediately.

Chapter Mix

Class 11 Mathematics: Sequences and Series

Related Formula

For finding extrema of a polynomial function f(x)$f(x)$, we set its derivative f'(x) = 0$f'(x) = 0$.

Core Logic

Given the 6th term of the A.P. is a₆ = 2$a_6 = 2$.

We need to maximize the product P = a₁ a₄ a₅$P = a_1 a_4 a_5$.

Substituting a = 2 - 5d$a = 2 - 5d$ into the expression for P$P$:

Step 1: Expand and Differentiate

Let's expand P$P$ as a function of d$d$, f(d)$f(d)$:

To find the maximum, we differentiate f(d)$f(d)$ with respect to d$d$ and equate to zero:

Factoring the quadratic:

This gives critical points d = (8)/(5)$d = \frac{8}{5}$ and d = (2)/(3)$d = \frac{2}{3}$.

Step 2: Check for Maximum

We check the second derivative to confirm a maximum:

At d = (8)/(5)$d = \frac{8}{5}$:

At d = (2)/(3)$d = \frac{2}{3}$:

Therefore, the greatest product occurs at d = (8)/(5)$d = \frac{8}{5}$.

Pattern Recognition

When asked to maximize a product of A.P. terms with a known constant term, express all terms strictly in d$d$, build the cubic, and use standard calculus f'(x)=0$f'(x)=0$ checking roots against the 2nd derivative test (Wavy Curve method works beautifully here).

Chapter Mix

Class 11 Mathematics: Sequences and Series Class 12 Mathematics: Application of Derivatives

Related Formula

Core Logic

For the first Geometric Progression (GP): First term t₁ = a$t_1 = a$ Third term t₃ = b = a r₁² ⇒ r₁² = (b)/(a)$t_3 = b = a r_1^2 \Rightarrow r_1^2 = \frac{b}{a}$ The 11th$11^{\text{th}}$ term is:

For the second Geometric Progression (GP): First term T₁ = a$T_1 = a$ Fifth term T₅ = a r₂⁴ = b ⇒ r₂⁴ = (b)/(a) ⇒ r₂ = ((b)/(a))1/4$T_5 = a r_2^4 = b \Rightarrow r_2^4 = \frac{b}{a} \Rightarrow r_2 = \left(\frac{b}{a}\right)^{1/4}$

Step 1: Equating the Terms

The pth$p^{\text{th}}$ term of the second GP is:

Given that t₁₁ = Tₚ$t_{11} = T_p$:

Step 2: Solving for p

Since a$a$ and b$b$ are distinct positive real numbers, (b)/(a) ≠ 1$\frac{b}{a} \neq 1$. Therefore, we can equate the exponents:

Pattern Recognition

Express the common ratios strictly in terms of powers of (b/a)$(b/a)$ to bypass isolated radical tracking.

Chapter Mix

Class 11 Maths: Sequences and Series