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From $232$ to $540$, find the number of multiples of $3$.
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Multiples of $3$ from $232$ to $540$ are $234, 237, 240, ......, 540$ (1 Mark)
This forms an AP with $a=234, d=3$ (1 Mark)
$540 = 234 + (n - 1)3$
$\Rightarrow n = 103$ (1 Mark)
This forms an AP with $a=234, d=3$ (1 Mark)
$540 = 234 + (n - 1)3$
$\Rightarrow n = 103$ (1 Mark)