Question
The sum of first 45 natural numbers is:
Answer: Option A
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Let Sn =(1 + 2 + 3 + ... + 45). This is an A.P. in which a =1, d =1, n = 45.
Sn = \(\frac{n}{2}\left[2a+(n-1)d\right]=\frac{45}{2}\times\left[2\times1+(45-1)\times1\right]=\left(\frac{45}{2}\times46\right)=45\times23\)
= 45 x (20 + 3)
= 45 x 20 + 45 x 3
= 900 + 135
=1035
Shorcut Method:
Sn = \(\frac{n(n+1)}{2} = \frac{45(45+1)}{2}=1035\)
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