JEE Advanced · Mathematics · 5. Sequences & Series
Let be consecutive terms of an arithmetic progression with common difference , and let , be consecutive terms of another arithmetic progression with common difference , where . For each , let be a rectangle with length , width and area .If , then the value of is _______.
- A 18900
- B 19000
- C 12450
- D 19124
Answer & Solution
Correct Answer
(A) 18900
Step-by-step Solution
Detailed explanation
Given \(l_1, l_2 \ldots l_{100}\) are consecutive terms of an \(A . P\)
Now let \(T_1=a\) and common difference \(=d_1\)
And similarly for \(A\). \(P w_1, w_2, \ldots w_{100}, T_1=b\) and common difference \(=d_2\)
Now given, \(A_{51}-A_{50}=l_{51} w_{51}-l_{50} w_{50}\)
\(\Rightarrow\left(a+50 d_1\right)\left(b+50 d_2\right)-\left(a+49 d_1\right)\) \(\left(b+49 d_2\right)=1000 \)
\( \Rightarrow 50 b d_1+50 a d_2+2500 d_1 d_2-49 a d_2~-\) \(49 b d_1-2401 d_1 d_2=1000 \)
\( \Rightarrow b d_1+a d_2+99 d_1 d_2=1000\)
So, \(b d_1+a d_2=10\left\{\right.\) as given \(\left.d_1 d_2=10\right\}\)
Now finding \(A_{100}-A_{90}=l_{100} w_{100}-l_{90} w_{90}\) we get,
\(=\left(a+99 d_1\right)\left(b+99 d_2\right)-\left(a+89 d_1\right)\) \(\left(b+89 d_2\right) \)
\( =99 b d_1+99 a d_2+99^2 d_1 d_2-89 b d_1\) \(-~89 a d_2-89^2 d_1 d_2 \)
\( =10\left(b d_1+a d_2\right)+1880 d_1 d_2 \)
\(=10(10)+18800 \text { \{again using } d_1 d_2=10\) & \(b d_1+a d_2=10\} \)
\( =18900\)
Now let \(T_1=a\) and common difference \(=d_1\)
And similarly for \(A\). \(P w_1, w_2, \ldots w_{100}, T_1=b\) and common difference \(=d_2\)
Now given, \(A_{51}-A_{50}=l_{51} w_{51}-l_{50} w_{50}\)
\(\Rightarrow\left(a+50 d_1\right)\left(b+50 d_2\right)-\left(a+49 d_1\right)\) \(\left(b+49 d_2\right)=1000 \)
\( \Rightarrow 50 b d_1+50 a d_2+2500 d_1 d_2-49 a d_2~-\) \(49 b d_1-2401 d_1 d_2=1000 \)
\( \Rightarrow b d_1+a d_2+99 d_1 d_2=1000\)
So, \(b d_1+a d_2=10\left\{\right.\) as given \(\left.d_1 d_2=10\right\}\)
Now finding \(A_{100}-A_{90}=l_{100} w_{100}-l_{90} w_{90}\) we get,
\(=\left(a+99 d_1\right)\left(b+99 d_2\right)-\left(a+89 d_1\right)\) \(\left(b+89 d_2\right) \)
\( =99 b d_1+99 a d_2+99^2 d_1 d_2-89 b d_1\) \(-~89 a d_2-89^2 d_1 d_2 \)
\( =10\left(b d_1+a d_2\right)+1880 d_1 d_2 \)
\(=10(10)+18800 \text { \{again using } d_1 d_2=10\) & \(b d_1+a d_2=10\} \)
\( =18900\)
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