Let l _ 1 , l _ 2 . . . l _ 100 be consecutive terms of an...

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, ${A}_{51} - {A}_{50} = {l}_{51}{w}_{51} - {l}_{50}{w}_{50}$

$⇒ \left(a + 50{d}_{1}\right)\left(b + 50{d}_{2}\right) - \left(a + 49{d}_{1}\right)\left(b + 49{d}_{2}\right) = 1000$

$⇒ 50b{d}_{1} + 50a{d}_{2} + 2500{d}_{1}{d}_{2} - 49a{d}_{2} - 49b{d}_{1} - 2401{d}_{1}{d}_{2} = 1000$

$⇒ b{d}_{1} + a{d}_{2} + 99{d}_{1}{d}_{2} = 1000$

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,

$= \left(a + 99{d}_{1}\right)\left(b + 99{d}_{2}\right) - \left(a + 89{d}_{1}\right)\left(b + 89{d}_{2}\right)$

$= 99b{d}_{1} + 99a{d}_{2} + {99}^{2}{d}_{1}{d}_{2} - 89b{d}_{1} - 89a{d}_{2} - {89}^{2}{d}_{1}{d}_{2}$

$= 10\left(b{d}_{1} + a{d}_{2}\right) + 1880{d}_{1}{d}_{2}$

$= 10\left(10\right) + 18800 \left\{\text{again using}{d}_{1}{d}_{2} = 10 \text{\text{ and } } b{d}_{1} + a{d}_{2} = 10\right\}$

$= 18900$