by 
aishwarya
ANSWERS: 6
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first term(a) = 1 common difference(d) = 2 n=100 sum of first n terms in an A.P. is given by n/2[2a + (n-1)d]
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1st odd number: 1 2nd odd number: 3 3rd odd number: 5 = 6 - 1 4th odd number: 7 = 8 - 1 The pattern is: nth odd number: [2*n - 1] Sum(nth odd number)= Sum(2*n - 1)= 2 * Sum(n) - Sum(1)= 2 * Sum(n) - n Moreover, we have: Sum(n)= n * (n+1)/2 If n is even, this can be shown if you group the numbers this way: (n+1) + ([n-1]+2) + ([n-2]+3) +... You get n/2 groups of value (n+1). Sum(nth odd number)= 2 * n * (n+1)/2 - n = n(n+1) -n = n*n For n=100, the sum is: 100 * 100 = 10,000 ---- ADDED ---- Here a simple trick to find the solution: "The first 100 odd numbers form a series: 1,3,5,7,9...191,193,195,197,199. To determine the sum of all of the numbers in this series, pair off the first and last number, etc., which yields (1 and 199), (3 and 197), (5 and 195)...(97 and 103) and (99 and 101). Notice that the sum of each pair is 200. Since there are 100 numbers, there are 50 pairs; therefore, the sum is 200 times 50, or 10,000." Source and further information: http://www.nsa.gov/kids/games/games00021.cfm
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the formula is n^2 =100*100=10000
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That would be the odd numbers from 1 to 199. Take the number in pairs, the smallest and largest (1 + 199), the next largest and next smallest (3 + 197) etc. Each pair will add up to 200. There will be 50 pairs. Your answer will then be the value of each pair multiplied by the number of pairs (200 * 50) This can also be represented as ((100*2)*(100/2)) which simplifies to (100*100), your original number of odd numbers squared. Which is 10000
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1 + 3 + 5 + 7 + 9 + ... + 97 + 99 Look at the first and last numbers: 1 + 99 Then the next two: 3 + 97 Then the next two 5 + 95 etc. 49 + 51 They all add up to 100 and you have 25 pairs. Answer: 2500
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first odd = 1 last odd = 199 sum of pair is 200 this holds equally from any pair (first+n)+(last-n) with 100 numbers to pair up we get 50 pairs. so the result is 50 * 200 = 10,000 regards JakobA
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