The sum of n terms of the series 2 , 5 , 8 , 11..... is 60100 then n is
Answers
Answered by
162
Solution:-
Given : a = 2 and d = 3 Sum = 60100.
Sn = n/2 {2a + (n - 1)d} = 60100
n/2 {4 + (n - 1)3} = 60100
n/2 {4 + 3n - 3} = 60100
n/2 {3n + n} = 60100
3n² + n = 60100 × 2
3n² + n - 120200 = 0
3n² + 601n - 600n - 120200 = 0
n(3n + 601) - 200 (3n + 601) = 0
(n - 200) (3n + 601) = 0
3n = - 601 or n = - 601/3
Here the value of n cannot be negative,
So, n = 200
Answer.
Given : a = 2 and d = 3 Sum = 60100.
Sn = n/2 {2a + (n - 1)d} = 60100
n/2 {4 + (n - 1)3} = 60100
n/2 {4 + 3n - 3} = 60100
n/2 {3n + n} = 60100
3n² + n = 60100 × 2
3n² + n - 120200 = 0
3n² + 601n - 600n - 120200 = 0
n(3n + 601) - 200 (3n + 601) = 0
(n - 200) (3n + 601) = 0
3n = - 601 or n = - 601/3
Here the value of n cannot be negative,
So, n = 200
Answer.
Answered by
65
a=2 d=3 sum=60100
Sum=n/2(2a+(n-1)d
60100=n/2 (4+(n-1)3)
12200=n(4+3n-3)
12200=n+3n2
3n2+n-12200
n=200. Ans
Hope it helps.....
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