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Find the sum of the first $40$ positive integers divisible by $6$.
A4680
B4920
C5100
D5280
Answer & Solution
Correct answer: B. 4920
The first 40 positive integers divisible by 6 are $6,12,18,\ldots,240$, an AP with $a=6$, $d=6$, and $n=40$. So
$$S_{40}=\frac{40}{2}(6+240)=20\cdot 246=4920.$$
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