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Question

The sum of how many terms of the series $$6+12+18+24+.....$$ is $$1800$$?


A
16
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B
24
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C
20
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D
18
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E
22
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Solution

The correct option is B $$24$$
This is an A.P in which $$a=6,d=6$$ and $${S}_{n}=1800$$
Then, $$\cfrac{n}{2}[2a+(n-1)d]=1800$$
$$\Rightarrow$$ $$\cfrac{n}{2}[2\times 6+(n-1)\times 6]=1800$$
$$\Rightarrow$$ $$3n(n+1)=1800$$
$$\Rightarrow$$ $$n(n+1)=600$$
$$\Rightarrow$$ $${n}^{2}+n-600=0$$
$$\Rightarrow$$$${n}^{2}+25n-24n-600=0$$
$$\Rightarrow$$ $$n(n+25)-24(n+25)=0$$
$$\Rightarrow$$ $$(n+5)(n-24)=0$$
$$\Rightarrow$$ $$n=24$$
Number of terms $$=24$$

Mathematics

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