Mathematics - Sequence & Series Question with Solution | TestHub

Given $l_1,l_2...l_{100}$ are consecutive terms of an $A.P$

Now let $T_1=a$ and common difference $=d_1$ 

And similarly for $A.P$ $w_1,w_2,...w_{100}$, $T_1=b$ and common difference $=d_2$

Now given, $A_{51}-A_{50}=l_{51}{w}_{51}-l_{50}{w}_{50}$

$\Rightarrow \left(a+50d_1\right)\left(b+50d_2\right)-\left(a+49d_1\right)\left(b+49d_2\right)=1000$

$\Rightarrow 50b{d}_1+50a{d}_2+2500d_1{d}_2-49a{d}_2-49b{d}_1-2401d_1{d}_2=1000$

$\Rightarrow b{d}_1+a{d}_2+99d_1{d}_2=1000$

So, $b{d}_1+a{d}_2=10$ $\left\{\text{as given }{d}_1{d}_2=10\right\}$

Now finding $A_{100}-A_{90}=l_{100}{w}_{100}-l_{90}{w}_{90}$ we get,

$=\left(a+99d_1\right)\left(b+99d_2\right)-\left(a+89d_1\right)\left(b+89d_2\right)$

$=99b{d}_1+99a{d}_2+{99}^2{d}_1{d}_2-89b{d}_1-89a{d}_2-{89}^2{d}_1{d}_2$

$=10\left(b{d}_1+a{d}_2\right)+1880d_1{d}_2$

$=10\left(10\right)+18800$ $\left\{\text{again using }{d}_1{d}_2=10 \& b{d}_1+a{d}_2=10\right\}$

$=18900$