arthham

JEE Advanced 2022 · Sequences and Series

Let l1,l2,…,l100l_1, l_2, \ldots, l_{100} be consecutive terms of an arithmetic progression with common difference d1d_1, and let w1,w2,…,w100w_1, w_2, \ldots, w_{100} be consecutive terms of another arithmetic progression with common difference d2d_2, where d1d2=10d_1 d_2 = 10. For each i=1,2,…,100i = 1, 2, \ldots, 100, let RiR_i be a rectangle with length lil_i, width wiw_i and area AiA_i. If A51−A50=1000A_{51} - A_{50} = 1000, then the value of A100−A90A_{100} - A_{90} is ______.

Answer: 1890018900

Watch this solved on the board The full walkthrough, written out a line at a time with narration explaining each step as it appears. Pause anywhere and ask why a line is there. Free to watch

Or read the worked solution below. Jump to the steps ↓

Written solution

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 d₁ d₂ = 10. For each i = 1, 2, …, 100, let Rᵢ be a rectangle with length lᵢ, width wᵢ and area Aᵢ. If A₅₁ - A₅₀ = 1000, then the value of A₁₀₀ - A₉₀ is ______.

Given: li=l1+(i−1)d1,wi=w1+(i−1)d2\text{Given: } l_i = l_1 + (i-1)d_1, \quad w_i = w_1 + (i-1)d_2

Ai=li⋅wiA_i = l_i \cdot w_i

d1d2=10d_1 d_2 = 10

To find: A100−A90, given A51−A50=1000\text{To find: } A_{100} - A_{90} \text{, given } A_{51} - A_{50} = 1000

Di=Ai+1−AiD_i = A_{i+1} - A_i

Ai+1=(li+d1)(wi+d2)A_{i+1} = (l_i + d_1)(w_i + d_2)

Di=lid2+wid1+d1d2D_i = l_i d_2 + w_i d_1 + d_1 d_2

Di+1−Di=d1d2+d2d1=2d1d2D_{i+1} - D_i = d_1 d_2 + d_2 d_1 = 2 d_1 d_2

Di+1−Di=2d1d2=20D_{i+1} - D_i = 2 d_1 d_2 = 20

D50=A51−A50=1000D_{50} = A_{51} - A_{50} = 1000

Di=1000+20(i−50)=20iD_i = 1000 + 20(i - 50) = 20i

A100−A90=∑i=9099DiA_{100} - A_{90} = \sum_{i=90}^{99} D_i

=∑i=909920i=20(90+91+⋯+99)= \sum_{i=90}^{99} 20i = 20(90 + 91 + \cdots + 99)

90+91+⋯+99=102(90+99)=5×189=94590 + 91 + \cdots + 99 = \dfrac{10}{2}(90 + 99) = 5 \times 189 = 945

A100−A90=20×945A_{100} - A_{90} = 20 \times 945

A100−A90=18900\boxed{A_{100} - A_{90} = 18900}

D50=20×50=1000 ✓D_{50} = 20 \times 50 = 1000 \ \checkmark

More from Sequences and Series

All 22 Sequences and Series questions →