Question 6 - Jee advanced Math 2022 P1 Questions with Solutions

Let \(l_{1}, l_{2},….,l_{100}\) be consecutive terms of an arithmetic progression with common difference \(d_{1}\), and let  \(w_{1}, w_{2},….,w_{100}\) be consecutive terms of another arithmetic progression with common difference \(d_{2}\), where \(d_{1}d_{2} = 10\). For each \(i = 1, 2,…,100\), let \(R_{i}\) be a rectangle with length \(l_{i}\), width \(w_{i}\) and area \(A_{i}\). If  \(A_{51}-A_{50}=1000\), then the value of  \(A_{100}-A_{90}\) is _____.

Sol : 

\(l_{1}, l_{2},….,l_{100}\) are consecutive terms of an arithmetic progression with common difference \(d_{1}\).
\(\implies l_{2} = l_{1} + d_{1}; \: l_{3} = l_{1} + 2 d_{1}; \: l_{4} = l_{1} + 3d_{1}; ….l_{i}= l_{1} + (i-1)d_{1}…\)
….where \(1 \leq i \leq 100\)

\(w_{1}, w_{2},….,w_{100}\) are consecutive terms of an arithmetic progression with common difference \(d_{2}\).
\(\implies w_{2} = w_{1} + d_{2}; \: w_{3} = w_{1} + 2 d_{2}; \: w_{4} = w_{1} + 3d_{2}; ….w_{j}= w_{1} + (j-1)d_{2}…\)
….where \(1 \leq j \leq 100\)

If \(A_{i}\) is the area of a rectangle \(R_{i}\), then the length and the width for that rectangle will be \(l_{i}\) and \(w_{i}\).

\(\implies A_{51}-A_{50} = l_{51}w_{51} - l_{50}w_{50}\)
                          \( = (l_{1}+50 d_{1}) (w_{1}+50 d_{2}) - (l_{1}+49 d_{1}) (w_{1}+49 d_{2})\)
                          \( =  l_{1}w_{1} + 50 l_{1}d_{2} + 50 w_{1}d_{1} + 50^{2} d_{1}d_{2} - [l_{1}w_{1} + 49 l_{1}d_{2} + 49 w_{1}d_{1} + (50-1)^{2} d_{1}d_{2}]\)

                         \( = l_{1}d_{2} + w_{1}d_{1} + (50^{2} - 50^{2} + 100 - 1)d_{1}d_{2}\)
                         \( = l_{1}d_{2} + w_{1}d_{1} + (99 \times 10)\)
                         \( = l_{1}d_{2} + w_{1}d_{1} + 990\)

\(\implies 1000 = l_{1}d_{2} + w_{1}d_{1} + 990\)
\(\implies l_{1}d_{2} + w_{1}d_{1} = 10\)

Consider,
\(A_{100}-A_{90} = l_{100}w_{100} - l_{90}w_{90}\)
                         \( = (l_{1}+99 d_{1}) (w_{1}+99 d_{2}) - (l_{1}+89 d_{1}) (w_{1}+89 d_{2})\)
                         \( =  l_{1}w_{1} + 99 l_{1}d_{2} + 99 w_{1}d_{1} + 99^{2} d_{1}d_{2} - [l_{1}w_{1} + 89 l_{1}d_{2} + 89 w_{1}d_{1} + (99-10)^{2} d_{1}d_{2}]\)

                         \( = 10l_{1}d_{2} + 10w_{1}d_{1} + (99^{2} - 99^{2} + 1880)d_{1}d_{2}\)
                         \( = 10(l_{1}d_{2} + w_{1}d_{1}) + (1880 \times 10)\)
                         \( = (10 \times 10) + 18800\)
                         \( = 18900\)

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