1+3+5+7...+199 find the sum
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AP:1,3,5,7.........199 Here a=1 d=2 an=199 As we know that;an=a+(n-1)d =1+(n-1)2=199 __n=100. Sn =n/2(a+an)__S100=100/2 (1+199)=50×200=100000
Sorry I have added 1 extra 0 at the end
Answers
Answered by
24
Hiii !
First term = a₁ = 1
Second term = a₂ = 3
Common difference = d = a₂- a₁ = 3 - 1 = 2
Last term = an = 199
an = a+ (n-1)d
199 = 1 + (n-1)2
198 = (n-1)2
198/2 = n - 1
99 = n - 1
n = 100
No: of terms = n = 100
Sum = Sn
Sn = n/2 [a + an]
= 100/2 [ 1 + 199]
= 50 [ 200]
= 10000
Sum = 10000
Hope this will help you
First term = a₁ = 1
Second term = a₂ = 3
Common difference = d = a₂- a₁ = 3 - 1 = 2
Last term = an = 199
an = a+ (n-1)d
199 = 1 + (n-1)2
198 = (n-1)2
198/2 = n - 1
99 = n - 1
n = 100
No: of terms = n = 100
Sum = Sn
Sn = n/2 [a + an]
= 100/2 [ 1 + 199]
= 50 [ 200]
= 10000
Sum = 10000
Hope this will help you
Answered by
9
10,000 is the answer
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