JEE Challenger
More from Sequences and Series
MathMedAdv 2022 • Paper 1

Area Difference of Rectangles Formed by Two Arithmetic Progressions

Let l1,l2,…,l100l_1, l_2, \dots, l_{100} be consecutive terms of an arithmetic progression with common difference d1d_1, and let w1,w2,…,w100w_1, w_2, \dots, 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, \dots, 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 ________.

Official Numerical Answer18900

Topics & Concepts

Sequences and SeriesArithmetic Progression

Step-by-Step Solution

To find the value of A100−A90A_{100} - A_{90}, we express the area AiA_i as Ai=liwi=(l1+(i−1)d1)(w1+(i−1)d2)A_i = l_i w_i = (l_1 + (i-1)d_1)(w_1 + (i-1)d_2).

Using the given condition A51−A50=1000A_{51} - A_{50} = 1000 alongside d1d2=10d_1 d_2 = 10, we evaluate the difference A51−A50=l1d2+w1d1+99d1d2=1000A_{51} - A_{50} = l_1 d_2 + w_1 d_1 + 99 d_1 d_2 = 1000, which yields l1d2+w1d1=10l_1 d_2 + w_1 d_1 = 10.

Finally, calculating A100−A90=10(l1d2+w1d1)+(992−892)d1d2=10(10)+1880(10)=18900A_{100} - A_{90} = 10(l_1 d_2 + w_1 d_1) + (99^2 - 89^2) d_1 d_2 = 10(10) + 1880(10) = 18900.

PreviousNext