how many terms of the progression 3,6,9,12... must be taken at the least to have a sum not less than 2000?
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Sum = n/2(2a+(n-1)d)
2000 = n/2(2*3+(n-1)3)
4000= n(6+3n-3)
4000= n(3+3n)
4000=3n+3nsquare
Then splitting middle term apply it answer will be two then the positive value will take...
Hope this might help u
2000 = n/2(2*3+(n-1)3)
4000= n(6+3n-3)
4000= n(3+3n)
4000=3n+3nsquare
Then splitting middle term apply it answer will be two then the positive value will take...
Hope this might help u
mamoni3:
i have solved up to here but from here i dont know how to solve... please help me to solve this problem
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