(51+52+53+.....+100 is equal to
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Which class question ???
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6
a = 51
d = 1
Tn = 100
=> a+ (n-1)d = 100
=> 51 + (n-1)(1) = 100
=> n-1 = 49
=> n = 50
S50 = 50/2 [ 2*51 + (50-1)(1)]
= 25 ( 102 + 49)
= 25 * 151
= 3775
Required sum = 3775
d = 1
Tn = 100
=> a+ (n-1)d = 100
=> 51 + (n-1)(1) = 100
=> n-1 = 49
=> n = 50
S50 = 50/2 [ 2*51 + (50-1)(1)]
= 25 ( 102 + 49)
= 25 * 151
= 3775
Required sum = 3775
Right
its correct but sorry
Why sorry ???
And It's a formula .... you will get the answer easily
kk thank
Which class yours Sir ???
Answered by
2
51 + 52 + 53 + 54 + 55 + 56 + 57 + 58 + 59 + 60 + 61 + 62 + 63 + 64 + 65 + 66 + 67 + 68 + 69 + 70 + 71 + 72 + 73 + 74 + 75 + 76 + 77 + 78 + 79 + 80 + 81 + 82 + 83 + 84 + 85 + 86 + 87 + 88 + 89 + 90 + 91 + 92 + 93 + 94 + 95 + 96 + 97 + 98 + 99 + 100 = 3775
its an easy method and funny thanks
No it's not easy .... .... if this question comes in your exam then what you can do ...
Will you start adding this numbers ???
no but generally kk sorry
Don't say Sorry
kk
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