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

For A.P.: a = b-d, c = b+d$a = b-d, c = b+d$. For G.P.: (2b²)² = a² c² 4b⁴ = a² c²$(2b^2)^2 = a^2 c^2 \implies 4b^4 = a^2 c^2$.

Core Logic

Given a + b + c = 1 (b-d) + b + (b+d) = 1 3b = 1 b = (1)/(3)$a + b + c = 1 \implies (b-d) + b + (b+d) = 1 \implies 3b = 1 \implies b = \frac{1}{3}$. Using G.P. condition:

Since a < b < c$a < b < c$, d > 0$d > 0$. Taking (1)/(9) - d² = -(2)/(9) d² = (1)/(3) d = 1√(3)$\frac{1}{9} - d^2 = -\frac{2}{9} \implies d^2 = \frac{1}{3} \implies d = \frac{1}{\sqrt{3}}$.

Step 1: Compute Required Expression

Pattern Recognition

Set a=b-d, c=b+d$a=b-d, c=b+d$ to eliminate terms early using a+b+c=1$a+b+c=1$.

Chapter Mix

Class 11 Maths: Sequences and Series

Related Formula

The n$n$-th term of a sequence is aₙ = Sₙ - Sₙ₋₁$a_n = S_n - S_{n-1}$.

Core Logic

We have Sₙ = α n² + β n$S_n = \alpha n^2 + \beta n$.

This shows the sequence is an Arithmetic Progression with common difference 2α$2\alpha$.

Step 1: Using Given Conditions

Condition 1: a₁₀ = 59$a_{10} = 59$

Condition 2: a₆ = 7a₁$a_6 = 7a_1$

Substitute into the condition:

Step 2: Solving for Alpha and Beta

Substitute 2α = 3β β = (2)/(3)α$2\alpha = 3\beta \implies \beta = \frac{2}{3}\alpha$ into the first equation:

If α = 3$\alpha = 3$, then β = (2)/(3)(3) = 2$\beta = \frac{2}{3}(3) = 2$. Therefore, α + β = 3 + 2 = 5$\alpha + \beta = 3 + 2 = 5$.

Pattern Recognition

Any sum of n$n$ terms that is purely a quadratic function in n$n$ describes an AP. You can instantly map the common difference to 2α$2\alpha$ and the first term to α + β$\alpha + \beta$.

Chapter Mix

Class 11 Maths: Sequences and Series

Related Formula

Sum of GP: Sₙ = a ( (1 - rⁿ)/(1 - r) )$S_n = a \left( \frac{1 - r^n}{1 - r} \right)$

Core Logic

The sequence is 3⁶, 3⁴, 3², 3⁰,$3^6, 3^4, 3^2, 3^0, \dots$ General term ak = 36 - 2(k-1) = 38 - 2k$a_k = 3^{6 - 2(k-1)} = 3^{8 - 2k}$.

The exponent is an AP with first term 6$6$, common difference -2$-2$, up to n$n$ terms. Sum = (n)/(2) [2(6) + (n-1)(-2)] = (n)/(2) [14 - 2n] = n(7 - n)$\frac{n}{2} [2(6) + (n-1)(-2)] = \frac{n}{2} [14 - 2n] = n(7 - n)$.

Step 1: Simplifying the Expression

Now we evaluate the sum:

Step 2: Geometric Series Sum

This is a GP with first term a = 3⁶$a = 3^6$, common ratio r = (1)/(3)$r = \frac{1}{3}$, and 40$40$ terms.

Step 3: Comparing with Given Form

Given expression:

Comparing with (3^α - 1)/(3^β)$\frac{3^\alpha - 1}{3^\beta}$, we get:

Check (40, 33) = 1$\gcd(40, 33) = 1$. (True) Therefore, α + β = 40 + 33 = 73$\alpha + \beta = 40 + 33 = 73$.

Pattern Recognition

Product sequences of exponent-based terms collapse neatly when finding the n$n$-th root, reducing complex product operations into standard Geometric Progressions.

Chapter Mix

Class 11 Maths: Sequences and Series

Related Formula

Core Logic

Given common difference d = -3/4$d = -3/4$. (n)/(2)(a₁ + (a₁)/(4)) = (525)/(2)$\frac{n}{2}\left(a_1 + \frac{a_1}{4}\right) = \frac{525}{2}$ (n)/(2) · (5a₁)/(4) = (525)/(2) ⇒ (5na₁)/(4) = 525 ⇒ n a₁ = 420$\frac{n}{2} \cdot \frac{5a_1}{4} = \frac{525}{2} \Rightarrow \frac{5na_1}{4} = 525 \Rightarrow n a_1 = 420$

Step 1: Finding Parameters

Use n$n$-th term formula:

Substitute into n a₁ = 420$n a_1 = 420$: n(n-1) = 420$n(n-1) = 420$

Since n > 0$n > 0$, n = 21$n = 21$. This means a₁ = 20$a_1 = 20$.

Step 2: Evaluating S17

Pattern Recognition

Dual AP conditions defining sum and last term invariably reduce to a quadratic in n$n$. Using Sₙ = (n)/(2)(a + l)$S_n = \frac{n}{2}(a + l)$ avoids messy expansion of (n-1)d$(n-1)d$ prematurely.

Chapter Mix

Class 11 Maths: Sequences and Series

Related Formula

Core Logic

Let the terms α₁, α₂, α₃, α₄$\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}$ be a - 3d, a - d, a + d, a + 3d$a - 3d, a - d, a + d, a + 3d$.

Sum of terms = 48$48$:

Step 1: Product Condition

We are given α₁α₂α₃α₄ + l⁴ = 361$\alpha_{1}\alpha_{2}\alpha_{3}\alpha_{4} + l^{4} = 361$. Since the common difference is l = 2d$l = 2d$, l⁴ = 16d⁴$l^4 = 16d^4$.

Substitute a = 12$a = 12$ (a² = 144$a^2 = 144$):

Step 2: Factoring the Quartic

Notice that 25d⁴ - 1440d² + 20736 = (5d² - 144)²$25d^4 - 1440d^2 + 20736 = (5d^2 - 144)^2$.

Case 1: 5d² = 144 + 19 = 163 d² = (163)/(5)$5d^2 = 144 + 19 = 163 \implies d^2 = \frac{163}{5}$ (Rejected, as d$d$ would not yield integer common differences).

Case 2: 5d² = 144 - 19 = 125 d² = 25 d = ± 5$5d^2 = 144 - 19 = 125 \implies d^2 = 25 \implies d = \pm 5$.

This gives common difference l = 2d = 10$l = 2d = 10$, which is an integer. Thus, d=5$d=5$ is valid.

Step 3: Calculating the Largest Term

The largest term of the A.P. is a + 3d$a + 3d$:

Pattern Recognition

The product of four symmetrically spaced A.P. terms (a-3d)(a-d)(a+d)(a+3d) + (2d)⁴$(a-3d)(a-d)(a+d)(a+3d) + (2d)^4$ strictly simplifies to a perfect square: (a² - 5d²)²$(a^2 - 5d^2)^2$. Knowing this identity instantly skips the heavy expansion algebra.

Chapter Mix

Class 11 Maths: Sequences and Series