The sum of the first 100 positive integers is 5050.What is the sum of the first 200 positive integers?
a. 10,100
b. 10,200
c. 15,050
d. 20,050
e. 20,100
The sum of the first
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Answer is E.
Sum of first 100 integers = 5050
Sum of first 200 integers would be (5050 *2) +(100 * 100) = 20100
C
E
The question states the sum of the first 200 positive integers i.e. 1+2+3+4…200
this can be solved with the formula: n(n+1)/2 –> 200 (200+1)/2 = 20100.
Can you please explain when do we use the above stated Formula……n(n+1)/2
@Manisha
this is the general formula for sum of n integers n(n+1)/2. Please study Arithmetic Progression for more details.
N(N+1)/2
= 200(200+1)/2
=20100
Ans E
Sum of n integers is given by n(n+1)/2
Here n=200
therefore, sum of 200 integeres =200(200+1)/2=20100
So correct answer is E
@ Thanku Bhaskar for telling me the usage of formula stated above.
Becoz its related to above question, i just wanted to know, if we have to find out the sum of integers from 101 to 200, do we have any formula for the same…….?
Since n(n+1)/2 can’t be used in such a case……
Answer E
using n(n+1)/2
Actually, the base formula is
S = n[2a + (n-1)d]/2
where n = number of terms, S = sum of terms, a = first term(=1)
and d = difference between the terms ( =1 )
so, our question is
1+2+3+4…………………+200
Just and FYI 200(l) is the last term term, so it can be calculated as,
l = a + (n-1)d
This can be look at like:
1 + 199 = 200
2 + 198 = 200
3 + 197 = 200
…
so there are 99 of those plus 1 more for the 200 value. We also have 1 100 value in the middle left so:
(200 * 100) + 100 = 20000 + 100 = 20100 (E)
Wow!!! Kamal that was a wonderful logic!!
c-is the correct answer
Ans : e
another logic :
5050 + (101+102+103….200)
5050 + ((100 +1)+(100+2) +100+3)…(100+100))
taking the 100 to 1 group and the rest to the other
5050 + ((100*100) + (1+2+3….100))
we no the 2nd group = 5050 and 100* 100 = 10000
5050 + (10000 + 5050) = 20100
thanks