Discussion :: Numbers
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The sum of first 45 natural numbers is
Answer : Option A
Explanation :
Let Sn =(1 + 2 + 3 + ... + 45). This is an A.P. in which a =1, d =1, n = 45.
sn=\( {n\over2}\)[2a + (n - 1)d]=\( {45\over 2}\)\(\times\)[2\(\times\)1+(45-1)\(\times\)1]=[\( {45 {} \over 2}\)\(\times\)46]=(45\(\times\)23)
= 45 x (20 + 3) |
= 45 x 20 + 45 x 3
= 900 + 135
= 1035.
Shorcut Method:
Sn = | n(n + 1) | = | 45(45 + 1) | = 1035. |
2 | 2 |
sn=\( {n(n+1) \ \over 2}\)=\( {45(45+1) \over 2}\)=1035
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