sum of how many terms of the arthmetic progression 3,7,11,....is 300?
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Sum of 25terms make 300
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a = 3, d = 4, Sn = 300
Sn = n/2[2a+(n-1)d]
n/2 [2*3 + (n-1)4] =300
n[ 6 + 4n - 4] = 600
6n + 4n^2 -4n = 600
4n^2 + 2n - 600 =0
I think you can factorize it further to get the answer.
Sn = n/2[2a+(n-1)d]
n/2 [2*3 + (n-1)4] =300
n[ 6 + 4n - 4] = 600
6n + 4n^2 -4n = 600
4n^2 + 2n - 600 =0
I think you can factorize it further to get the answer.
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