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jee solver Grade: 12th pass
        
wat is 1+3+5+7+9+11+......99 =  ?
how to do it fast
2 years ago

Answers : (6)

Prajwal Kavad
603 Points
										
there are total 50 consicutive odd numbers strating from 1 therefore  sum is (number of terms)^2 ,hence answer is 50^2=2500 is the answer
2 years ago
A M S ARUN KRISHNA
217 Points
										
lets add the first number and the last number
1 + 99 = 100
lets add the second number and the second last number
3 + 97 = 100
like that it goes on
5 + 95 = 100
7 + 93 = 100
9 + 91 = 100......
49 + 51 = 100
like this there are twenty five set of numbers
 so 25 × 100 = 2500
2500 is the answer
 
2 years ago
Prajwal Kavad
603 Points
										
wow what an answer by @arun krishna
2 years ago
V. Harshit -
84 Points
										
this answer can be solved by arithmetic progression..where a = 1, an=99 n=50 Sn=n/2(a+an)=25(100)=2500..
2 years ago
vedant patil
40 Points
										
for such type of questions add first term and last term then multipy with total then divide by 2 for ur case {1+99}*50/2 and u will get the answer.
2 years ago
Aakash
110 Points
										
Sum of n odd terms is ^{n^{2}}
2 years ago
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