Let a₁, a₂, …, a₂₀₂₄$a_1, a_2, \ldots, a_{2024}$ be an Arithmetic Progression such that a₁ + (a₅ + a₁₀ + a₁₅ + … + a₂₀₂₀) + a₂₀₂₄ = 2233$a_1 + (a_5 + a_{10} + a_{15} + \ldots + a_{2020}) + a_{2024} = 2233$. Then a₁ + a₂ + a₃ + … + a₂₀₂₄$a_1 + a_2 + a_3 + \ldots + a_{2024}$ is equal to

Related Formula

Core Logic

Let the terms of the GP be a, ar, ar², ar³,$a, ar, ar^2, ar^3, \dots$ Given: a₂ · a₃ · a₄ = 64$a_2 \cdot a_3 \cdot a_4 = 64$

Since it is a GP of increasing positive terms, r > 1$r > 1$ and a > 0$a > 0$.

Step 1: Utilize the sum condition

Given: a₁ + a₃ + a₅ = (813)/(7)$a_1 + a_3 + a_5 = \frac{813}{7}$

Extract a$a$ from ar² = 4 ⇒ a = (4)/(r²)$ar^2 = 4 \Rightarrow a = \frac{4}{r^2}$. Substitute this in:

Let r² = t$r^2 = t$:

Wait, there is a much faster method by just scaling the required expression.

Step 2: Calculate the required expression

We need a₃ + a₅ + a₇ = ar² + ar⁴ + ar⁶$a_3 + a_5 + a_7 = ar^2 + ar^4 + ar^6$. Notice that ar² + ar⁴ + ar⁶ = r² (a + ar² + ar⁴)$ar^2 + ar^4 + ar^6 = r^2 (a + ar^2 + ar^4)$. So, required sum = r² ((813)/(7))$r^2 \left(\frac{813}{7}\right)$.

To find r²$r^2$, we solve the quadratic in t = r²$t = r^2$:

The roots are t = 28$t = 28$ and t = (1)/(28)$t = \frac{1}{28}$. Since the GP is increasing, r > 1 ⇒ r² = 28$r > 1 \Rightarrow r^2 = 28$.

Step 3: Final evaluation

Alternatively, expand directly:

(Note: r² × (813)/(7) = 28 × (813)/(7) = 4 × 813 = 3252$r^2 \times \frac{813}{7} = 28 \times \frac{813}{7} = 4 \times 813 = 3252$)

Pattern Recognition

In GP questions demanding a sum shifted by a fixed index (like a₁+a₃+a₅$a_1+a_3+a_5$ to a₃+a₅+a₇$a_3+a_5+a_7$), immediately look to factor out the common ratio multiplier r^k$r^k$. Here it's a simple scaling by r²$r^2$.

Chapter Mix

Class 11 Maths: Sequences and Series Class 10 Maths: Quadratic Equations

Related Formula

Partial fraction decomposition for telescopic summing:

Core Logic

Given recurrence:

Rewrite the numerator to split the fraction:

Step 1: Telescope generation

Multiply both sides by appropriate powers of 2 to create a cancelling chain: For n=1$n=1$: a₂ - (1)/(2)a₁ = (2)/(2²) - (1)/(1²)$a_2 - \frac{1}{2}a_1 = \frac{2}{2^2} - \frac{1}{1^2}$ For n=2$n=2$: multiply by 2 ⇒ 2[a₃ - (1)/(2)a₂ = (2)/(3²) - (1)/(2²)] ⇒ 2a₃ - a₂ = (2 × 2)/(3²) - (2)/(2²)$\Rightarrow 2\left[a_3 - \frac{1}{2}a_2 = \frac{2}{3^2} - \frac{1}{2^2}\right] \Rightarrow 2a_3 - a_2 = \frac{2 \times 2}{3^2} - \frac{2}{2^2}$ Wait, let's look at a cleaner telescopic scaling: aₙ₊₁ - (2)/((n+1)²) = (1)/(2) (aₙ - (2)/(n²))$a_{n+1} - \frac{2}{(n+1)^2} = \frac{1}{2} \left(a_n - \frac{2}{n^2}\right)$.

Step 2: Identify Geometric Progression

Let Vₙ = aₙ - (2)/(n²)$V_n = a_n - \frac{2}{n^2}$. The recurrence gives Vₙ₊₁ = (1)/(2) Vₙ$V_{n+1} = \frac{1}{2} V_n$. This proves Vₙ$V_n$ is a geometric progression with common ratio r = 1/2$r = 1/2$. First term V₁ = a₁ - (2)/(1²) = 1 - 2 = -1$V_1 = a_1 - \frac{2}{1^2} = 1 - 2 = -1$.

Step 3: Infinite Summation

We need | Σn=1∞ ( aₙ - (2)/(n²) ) | = | Σn=1∞ Vₙ |$\left| \sum_{n=1}^{\infty} \left( a_n - \frac{2}{n^2} \right) \right| = \left| \sum_{n=1}^{\infty} V_n \right|$. Since Vₙ$V_n$ is an infinite GP:

Taking absolute value: |-2| = 2$|-2| = 2$

Pattern Recognition

When dealing with rational fraction recurrences Aₙ₊₁ - k Aₙ = f(n) - k f(n-1)$A_{n+1} - k A_n = f(n) - k f(n-1)$, immediately substitute Vₙ = Aₙ - f(n)$V_n = A_n - f(n)$. This substitution instantly isolates a classical Geometric Progression.

Chapter Mix

Class 11 Maths: Sequences and Series

Related Formula

Core Logic

The given sequence is a G.P. with ratio 1√(2)$\frac{1}{\sqrt{2}}$.

Thus, a₁, a₂, a₃, , a₁₀$a_1, a_2, a_3, \dots, a_{10}$ forms a standard G.P. with first term a₁$a_1$ and common ratio R = √(2)$R = \sqrt{2}$.

Step 1: Calculate the Sum

Sum of this new sequence is S₁₀ = 62$S_{10} = 62$.

Since (√(2))¹⁰ = 2⁵ = 32$(\sqrt{2})^{10} = 2^5 = 32$:

Pattern Recognition

If a sequence bₙ = aₙkⁿ⁻¹$b_n = \frac{a_n}{k^{n-1}}$ is a G.P. with ratio r$r$, then the base sequence aₙ$a_n$ is inherently a G.P. with ratio R = kr$R = kr$.

Chapter Mix

Class 11 Maths: Sequence and Series

Related Formula

Core Logic

Given the sum of first 4 terms is S₄ = 6$S_4 = 6$:

Given the sum of first 6 terms is S₆ = 4$S_6 = 4$:

Step 1: Finding Parameters a and d

Subtract equation (1) from equation (2):

Substitute d$d$ into equation (1):

Step 2: Calculating Sum of 12 terms

Pattern Recognition

Standard two-variable linear equations format strictly from sum identities. Fractions easily resolve by distributing the external multiplication constant n/2$n/2$ directly into the bracket.

Chapter Mix

Class 11 Maths: Sequences and Series

Related Formula

Exponential functional equation: f(x+y) = f(x)f(y) f(x) = a^x$f(x+y) = f(x)f(y) \implies f(x) = a^x$. Geometric progression sum formula: Sₙ = (a(rⁿ - 1))/(r - 1)$S_n = \frac{a(r^n - 1)}{r - 1}$.

Core Logic

Since f(1) = 7$f(1) = 7$, f(x) = 7^x$f(x) = 7^x$. For g(x+y) = g(xy)$g(x+y) = g(xy)$, set y = 1 g(x+1) = g(x)$y = 1 \implies g(x+1) = g(x)$. Since g(1) = 1$g(1) = 1$, we have g(1) = g(2) = = g(n) = 1$g(1) = g(2) = \dots = g(n) = 1$.

Step 1: Solve Summation

Since 7⁵ = 16807$7^5 = 16807$, n = 5$n = 5$.

Pattern Recognition

Recognize f(x)=7^x$f(x)=7^x$ and constant function g(x)=1$g(x)=1$ from given functional equations.

Chapter Mix

Class 11 Maths: Sequences and Series Class 11 Maths: Functions and Graphs