Aptitude - Numbers - Discussion
Discussion Forum : Numbers - General Questions (Q.No. 20)
20.
The sum of first 45 natural numbers is:
Answer: Option
Explanation:
Let Sn =(1 + 2 + 3 + ... + 45). This is an A.P. in which a =1, d =1, n = 45.
| Sn = | n | [2a + (n - 1)d] | = | 45 | x [2 x 1 + (45 - 1) x 1] | = |  |
45 | x 46 |  |
= (45 x 23) |
| 2 | 2 | 2 |
= 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 |
Discussion:
19 comments Page 2 of 2.
Tohid said:
8 years ago
n(n + 1)÷2 = 45(45 + 1)÷2 = 1035.
(1)
Piyosh said:
10 years ago
Is this n(n+1)/2 applicable for all A.P?
Md Ikramul Haque said:
7 years ago
4+5=9.
And 1035=1+0+3+5=9.
Answer is A.
And 1035=1+0+3+5=9.
Answer is A.
(13)
Bhartari Shinde said:
6 years ago
n(n+1)/2 = 45(45+1)/2 = 2070/2 = 1035.
(1)
Bill said:
1 decade ago
Help me to find even more easy way.
Ninjahari said:
6 years ago
n(n+1)/2 is applicable. Am I right?
(1)
Rinkal said:
8 years ago
I like your method, thanks @Rohit.
Puneet said:
1 decade ago
[N/2]*(first term + last term).
Jignesh said:
8 years ago
Good method @Rohit.
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