Question

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

A
16
B
24
C
20
D
18
E
22

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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