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← JEE Advanced 2022 · Paper 1
MathematicsNumerical+3Question 6 of 54

JEE Advanced 2022 · Paper 1, Mathematics question 6

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_1d_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 and worked solution

Official answer: 18900

First differences. Let Dn=An+1−AnD_n=A_{n+1}-A_n. Using ln+1=ln+d1l_{n+1}=l_n+d_1 and wn+1=wn+d2w_{n+1}=w_n+d_2:
Dn=lnd2+wnd1+d1d2.D_n=l_nd_2+w_nd_1+d_1d_2.

They form an AP. lnl_n and wnw_n grow by d1d_1 and d2d_2, so
Dn+1−Dn=2d1d2=20.D_{n+1}-D_n=2d_1d_2=20.
With D50=1000D_{50}=1000, Dn=1000+20(n−50)D_n=1000+20(n-50).

Sum.
A100−A90=∑n=9099Dn=10×1000+20∑n=9099(n−50)=10000+20×445=18900.A_{100}-A_{90}=\sum_{n=90}^{99}D_n=10\times1000+20\sum_{n=90}^{99}(n-50)=10000+20\times445=18900.