IIT-JEE Advanced 2022 Maths Paper 1 Problem 6

Problem

Let l₁, l₂, … , l₁₀₀ be consecutive terms of an arithmetic progression with common difference d₁, and let w₁, w₂, … , w₁₀₀ be consecutive terms of another arithmetic progression with common difference d₂, where d₁d₂ = 10. For each i = 1, 2, …, 100, let R_i be a rectangle with length l_i, width w_i and area A_i. If A₅₁ − A₅₀ = 1000, then find the value of A₁₀₀ − A₉₀.

Solution

Let’s start with the last equation, i.e., A₅₁ − A₅₀ = 1000. We’ll try to express this in terms of l_i and w_i, and see if we can use that expression to find A₁₀₀ − A₉₀.

In terms of the length and width, this equation can be rewritten as l₅₁w₅₁ – l₅₀w₅₀ = 1000.

Since l₅₁ = l₅₀ + d₁ and w₅₁ = w₅₀ + d₂, the previous equation becomes (l₅₀ + d₁)(w₅₀ + d₂) – l₅₀w₅₀ = 1000.

On expanding the first product, we’ll be left with l₅₀d₂ + w₅₀d₁ + d₁d₂ = 1000. Let’s call this equation I.

Now, let’s take a look at A₁₀₀ − A₉₀. This can be written as l₁₀₀w₁₀₀ – l₉₀w₉₀ or (l₉₀ + 10d₁)(w₉₀ + 10d₂) – l₉₀w₉₀.

If we simplify this, we’ll get 10l₉₀d₂ + 10w₉₀d₁ + 100d₁d₂, or 10(l₉₀d₂ + w₉₀d₁ + 10d₁d₂).

Now, using the various AP formulas, we could express l₉₀ and w₉₀ in terms of l₅₀ and w₅₀ respectively, and hope that we the terms from equation I.

But we can do slightly better. Let’s assume the value of the expression to be N.

Then, l₉₀d₂ + w₉₀d₁ + 10d₁d₂ = N/10. Let’s call this equation II.

If we subtract equation I from II, we get d₂(l₉₀ – l₅₀) + d₁(w₉₀ – w₅₀) + 9d₁d₂ = N/10 – 1000

Now, l₉₀ – l₅₀ = 40d₁ and w₉₀ – w₅₀ = 40d₂. So, previous equation changes to 40d₁d₂ + 40d₁d₂ + 9d₁d₂ = N/10 – 1000

So, N equals 10(1000 + 89d₁d₂). Finally, using d₁d₂ = 10, the value of N simplifies to 18900.

Comments

Too many variables (l_i, w_i, R_i, A_i, d₁, d₂) to scare you away. But beyond that, nothing too difficult. If you dont give up, and dont make any calculation mistakes along the way, you’ll surely reach your destination.