JEE Advanced 2022 Paper 1 · Q06 · Sequences & Series

Question

Let $l_{1},l_{2},\ldots,l_{100}$ be consecutive terms of an arithmetic progression with common difference $d_{1}$, and let $w_{1},w_{2},\ldots,w_{100}$ be consecutive terms of another arithmetic progression with common difference $d_{2}$, where $d_{1}d_{2}=10$. For each $i=1,2,\ldots,100$, 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 ____________.

SolveKar AI · Accepted Answer

$A_{i}=l_{i}w_{i}$. Then $A_{i+1}-A_{i}=l_{i+1}w_{i+1}-l_{i}w_{i}=l_{i+1}w_{i+1}-l_{i+1}w_{i}+l_{i+1}w_{i}-l_{i}w_{i}=l_{i+1}d_{2}+w_{i}d_{1}=(l_{i}+d_{1})d_{2}+w_{i}d_{1}=l_{i}d_{2}+w_{i}d_{1}+d_{1}d_{2}$. So the sequence $b_{i}=A_{i+1}-A_{i}$ is itself an AP with common difference $d_{1}d_{2}+d_{1}d_{2}=2d_{1}d_{2}=20$ (since each step adds $d_{1}d_{2}$ from the $l$ term and $d_{1}d_{2}$ from the $w$ term). Given $b_{50}=A_{51}-A_{50}=1000$. So $b_{n}=1000+(n-50)\cdot 20$. Now $A_{100}-A_{90}=\sum_{n=90}^{99}b_{n}=\sum_{n=90}^{99}[1000+20(n-50)]=10\cdot 1000+20\sum_{n=90}^{99}(n-50)=10000+20\cdot(40+41+\cdots+49)=10000+20\cdot 445=10000+8900=18900$.

JEE Advanced 2022 Paper 1 · Q06 · Sequences & Series

Solution